poj 3352 Road Construction && poj 3177 Redundant Paths
给一个无向图，问最少需要添加多少条边，使它成为双连通图。
做法：边双连通缩点，成为一棵树。若要使得任意一棵树，变成一个双连通图，那么至少增加的边数 =（度数为1的结点数 + 1 ）/ 2
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<set>
9 #include<map>
10 #include<queue>
11 #include<bitset>
12
13 using namespace std;
14
15
16 #pragma comment(linker, "/STACK:1024000000,1024000000")
17 #define inf 0x3f3f3f3f
18 #define eps 1e-9
19 #define FOR(i,s,t) for(int i = s; i < t; ++i )
20 #define REP(i,s,t) for( int i = s; i <= t; ++i )
21 #define pii pair<int,int>
22 #define MP make_pair
23 #define ls i << 1
24 #define rs ls | 1
25 #define md ((ll + rr) >> 1)
26 #define lson ll, md, ls
27 #define rson md + 1, rr, rs
28 #define LL long long
29 #define N 1010
30 #define M 2020
31 #define Pi acos(-1.0)
32 #define mod 1000007
33 #define ULL unsigned long long
34
35 int fst[N], nxt[M], vv[M], e;
36 void init(){
37 memset(fst, -1, sizeof fst); e = 0;
38 }
39 void add(int u, int v){
40 vv[e] = v, nxt[e] = fst[u], fst[u] = e++;
41 }
42
43 int bccno[N], pre[N], dc, bcnt;
44 bool cut[M];
45 int dfs(int u, int p){
46 int lowu = pre[u] = ++dc;
47 for(int i = fst[u]; ~i; i = nxt[i]){
48 int v = vv[i];
49 if(!pre[v]){
50 int lowv = dfs(v, u);
51 lowu = min(lowu, lowv);
52 if(lowv > pre[u]){
53 cut[i] = cut[i^1] = 1;
54 }
55 }
56 else if(v != p)
57 lowu = min(lowu, pre[v]);
58 }
59 return lowu;
60 }
61 void dfs2(int u, int cnt){
62 bccno[u] = cnt;
63 for(int i = fst[u]; ~i; i = nxt[i]){
64 int v = vv[i];
65 if(!bccno[v] && cut[i] == 0)
66 dfs2(v, cnt);
67 }
68 }
69 int in[N];
70 int main(){
71 int n, m;
72 while(scanf("%d%d", &n, &m) != EOF){
73 init();
74 for(int i = 0; i < m; ++i){
75 int u, v;
76 scanf("%d%d", &u, &v);
77 add(u, v), add(v, u);
78 }
79 dc = 0;
80 memset(pre, 0, sizeof pre);
81 memset(cut, 0, sizeof cut);
82 dfs(1, -1);
83 bcnt = 0;
84 memset(bccno, 0, sizeof bccno);
85 for(int i = 1; i <= n; ++i){
86 if(!bccno[i])
87 dfs2(i, ++bcnt);
88 }
89 memset(in, 0, sizeof in);
90 for(int i = 1; i <= n; ++i){
91 for(int j = fst[i]; ~j; j = nxt[j]){
92 int v = vv[j];
93 if(bccno[i] == bccno[v]) continue;
94 in[bccno[i]]++, in[bccno[v]]++;
95 }
96 }
97 int tot = 0;
98 for(int i = 1; i <= bcnt; ++i){
99 if(in[i] == 2) tot++;
100 }
101 printf("%d\n", (tot+1)/2);
102 }
103 return 0;
104 }
hdu 4738 Caocao‘s Bridges
有n座岛和m条桥，每条桥上有w个兵守着，现在要派不少于守桥的士兵数的人去炸桥，只能炸一条桥，使得这n座岛不连通，求最少要派多少人去。
做法：找权值最小的桥。有几个坑点：图不连通，输出0；有重边；桥上的士兵数为0，输出1。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<set>
9 #include<map>
10 #include<queue>
11 #include<bitset>
12
13 using namespace std;
14
15
16 #pragma comment(linker, "/STACK:1024000000,1024000000")
17 #define inf 0x3f3f3f3f
18 #define eps 1e-9
19 #define FOR(i,s,t) for(int i = s; i < t; ++i )
20 #define REP(i,s,t) for( int i = s; i <= t; ++i )
21 #define pii pair<int,int>
22 #define MP make_pair
23 #define ls i << 1
24 #define rs ls | 1
25 #define md ((ll + rr) >> 1)
26 #define lson ll, md, ls
27 #define rson md + 1, rr, rs
28 #define LL long long
29 #define N 1010
30 #define M 2000020
31 #define Pi acos(-1.0)
32 #define mod 1000007
33 #define ULL unsigned long long
34
35 int fst[N], nxt[M], vv[M], cost[M], uu[M], num[N][N], e;
36 void init(){
37 memset(num, 0, sizeof num);
38 memset(fst, -1, sizeof fst); e = 0;
39 }
40 void add(int u, int v, int c){
41 uu[e] = u, vv[e] = v, nxt[e] = fst[u], cost[e] = c, fst[u] = e++;
42 }
43
44 int pre[N], bccno[N], dc, bcnt;
45 bool cut[M];
46 int dfs(int u, int p){
47 int lowu = pre[u] = ++dc;
48 for(int i = fst[u]; ~i; i = nxt[i]){
49 int v = vv[i];
50 if(!pre[v]){
51 int lowv = dfs(v, u);
52 lowu = min(lowu, lowv);
53 if(lowv > pre[u])
54 cut[i] = cut[i^1] = 1;
55 }
56 else if(v != p)
57 lowu = min(lowu, pre[v]);
58 }
59 return lowu;
60 }
61 void dfs2(int u, int cnt){
62 bccno[u] = cnt;
63 for(int i = fst[u]; ~i; i = nxt[i]){
64 int v = vv[i];
65 if(!bccno[v] &&cut[i] == 0)
66 dfs2(v, cnt);
67 }
68 }
69 int main(){
70 int n, m;
71 while(scanf("%d%d", &n, &m) != EOF){
72 if(!n && !m) break;
73 init();
74 for(int i = 0; i < m; ++i){
75 int u, v, c;
76 scanf("%d%d%d", &u, &v, &c);
77 add(u, v, c);
78 add(v, u, c);
79 num[u][v]++, num[v][u]++;
80 }
81 dc = 0;
82 memset(cut, 0, sizeof cut);
83 memset(pre, 0, sizeof pre);
84 dfs(1, -1);
85 bool ok = 0;
86 for(int i = 1; i <= n; ++i){
87 if(!pre[i]) ok = 1;
88 }
89 if(ok){
90 puts("0"); continue;
91 }
92 bcnt = 0;
93 memset(bccno, 0, sizeof bccno);
94 for(int i = 1; i <= n; ++i){
95 if(!bccno[i])
96 dfs2(i, ++bcnt);
97 }
98 //printf("bcnt %d\n", bcnt);
99 if(bcnt == 1){
100 puts("-1"); continue;
101 }
102 int ans = inf;
103 for(int i = 0; i < e; i += 2){
104 if(cut[i] && num[uu[i]][vv[i]] == 1)
105 ans = min(ans, cost[i]);
106 }
107 if(ans == 0) ans++;
108 printf("%d\n", ans);
109 }
110 return 0;
111 }
poj 2942 Knights of the Round Table
圆桌骑士。大白的题目。
做法：点双连通。对每个连通分量判断是不是二分图。因为二分图是没有奇圈的。如果不能够染成二分图，则有奇圈，这个连通分量的人都是能够出席的。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<vector>
9 #include<stack>
10 #include<queue>
11 #include<bitset>
12 using namespace std;
13
14
15 #pragma comment(linker, "/STACK:1024000000,1024000000")
16 #define inf 0x3f3f3f3f
17 #define eps 1e-12
18 #define FOR(i,s,t) for(int i = s; i < t; ++i )
19 #define REP(i,s,t) for( int i = s; i <= t; ++i )
20 #define pii pair<int,int>
21 #define MP make_pair
22 #define ls i << 1
23 #define rs ls | 1
24 #define md ((ll + rr) >> 1)
25 #define lson ll, md, ls
26 #define rson md + 1, rr, rs
27 #define LL long long
28 #define N 1020
29 #define M 2000020
30 #define Pi acos(-1.0)
31 #define mod 1000000007
32
33 int fst[N], nxt[M], vv[M], e;
34 void init(){
35 memset(fst, -1, sizeof fst); e = 0;
36 }
37 void add(int u, int v){
38 vv[e] = v, nxt[e] = fst[u], fst[u] = e++;
39 }
40
41 struct edge{
42 int u, v;
43 edge(int u = 0, int v = 0) : u(u), v(v) {}
44 };
45 stack<edge> E;
46 vector<int> dcc[N];
47 int pre[N], dccno[N], dc, dcnt;
48 int dfs(int u, int p){
49 int lowu = pre[u] = ++dc;
50 for(int i = fst[u]; ~i; i = nxt[i]){
51 int v = vv[i];
52 edge e(u, v);
53 if(!pre[v]){
54 E.push(e);
55 int lowv = dfs(v, u);
56 lowu = min(lowv, lowu);
57 if(lowv >= pre[u]){
58 dcnt++;
59 dcc[dcnt].clear();
60 while(1){
61 edge x = E.top(); E.pop();
62 if(dccno[x.u] != dcnt){
63 dcc[dcnt].push_back(x.u);
64 dccno[x.u] = dcnt;
65 }
66 if(dccno[x.v] != dcnt){
67 dcc[dcnt].push_back(x.v);
68 dccno[x.v] = dcnt;
69 }
70 if(x.u == u && x.v == v) break;
71 }
72 }
73 }
74 else if(pre[v] < pre[u] && v != p)
75 lowu = min(lowu, pre[v]);
76 }
77 return lowu;
78 }
79
80 int col[N], odd[N];
81 bool bipartite(int u, int p, int b){
82 for(int i = fst[u]; ~i; i = nxt[i]){
83 int v = vv[i];
84 if(v == p || dccno[v] != b) continue;
85 if(col[v] == col[u]) return 0;
86 if(!col[v]){
87 col[v] = 3 - col[u];
88 if(!bipartite(v, u, b)) return 0;
89 }
90 }
91 return 1;
92 }
93 bool g[N][N];
94 int main(){
95 int n, m;
96 while(scanf("%d%d", &n, &m) != EOF){
97 if(!n && !m) break;
98 memset(g, 0, sizeof g);
99 for(int i = 0; i < m; ++i){
100 int u, v;
101 scanf("%d%d", &u, &v);
102 g[u][v] = g[v][u] = 1;
103 }
104 init();
105 for(int i = 1; i <= n; ++i){
106 for(int j = i+1; j <= n; ++j){
107 if(!g[i][j]){
108 add(i, j), add(j, i);
109 g[i][j] = g[j][i] = 1;
110 }
111 }
112 }
113 dc = dcnt = 0;
114 memset(pre, 0, sizeof pre);
115 memset(dccno, 0, sizeof dccno);
116 for(int i = 1; i <= n; ++i){
117 if(!pre[i]) dfs(i, -1);
118 }
119 // printf("dcnt %d\n", dcnt);
120 memset(odd, 0, sizeof odd);
121 for(int i = 1; i <= dcnt; ++i){
122 for(int j = 0; j < dcc[i].size(); ++j) dccno[dcc[i][j]] = i;
123 memset(col, 0, sizeof col);
124 int u = dcc[i][0];
125 col[u] = 1;
126 if(!bipartite(u, -1, i))
127 for(int j = 0; j < dcc[i].size(); ++j)
128 odd[dcc[i][j]] = 1;
129 }
130 int ans = n;
131 for(int i = 1; i <= n; ++i)
132 if(odd[i]) ans--;
133 printf("%d\n", ans);
134 }
135 }
hdu 3394 Railway
题意：一个无向图，有 n 个景点，如果至少两个环公用一条路，路上的游客就会发生冲突；如果一条路不属于任何的环，这条路就没必要修。问，有多少路不必修，有多少路会发生冲突
做法：没必要修的路就是桥。会发生冲突，则点双连通后，若连通分量里的边数大于点数，则这个连通分量里的路都是冲突边。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<stack>
9 #include<vector>
10 #include<queue>
11 #include<bitset>
12
13 using namespace std;
14
15
16 #pragma comment(linker, "/STACK:1024000000,1024000000")
17 #define inf 0x3f3f3f3f
18 #define eps 1e-9
19 #define FOR(i,s,t) for(int i = s; i < t; ++i )
20 #define REP(i,s,t) for( int i = s; i <= t; ++i )
21 #define pii pair<int,int>
22 #define MP make_pair
23 #define ls i << 1
24 #define rs ls | 1
25 #define md ((ll + rr) >> 1)
26 #define lson ll, md, ls
27 #define rson md + 1, rr, rs
28 #define LL long long
29 #define N 10010
30 #define M 200020
31 #define Pi acos(-1.0)
32 #define mod 1000007
33 #define ULL unsigned long long
34
35 int fst[N], vv[M], nxt[M], e;
36 void init(){
37 memset(fst, -1, sizeof fst); e = 0;
38 }
39 void add(int u, int v){
40 vv[e] = v, nxt[e] = fst[u], fst[u] = e++;
41 }
42
43
44 struct edge{
45 int u, v;
46 edge(int u = 0, int v = 0) : u(u), v(v) {}
47 };
48 stack<edge> E;
49 vector<int> dcc[N];
50 int dccno[N], pre[N], dc, dcnt, ans1, ans2;
51 bool in[N];
52 void calc(vector<int> &g){
53 memset(in, 0, sizeof in);
54 for(int i = 0; i < g.size(); ++i)
55 in[g[i]] = 1;
56 int tmp = 0;
57 for(int i = 0; i < g.size(); ++i){
58 int u = g[i];
59 for(int j = fst[u]; ~j; j = nxt[j]){
60 int v = vv[j];
61 if(in[v])
62 tmp++;
63 }
64 }
65 tmp /= 2;
66 if(tmp > g.size())
67 ans2 += tmp;
68 }
69 int dfs(int u, int p){
70 int lowu = pre[u] = ++dc;
71 for(int i = fst[u]; ~i; i = nxt[i]){
72 int v = vv[i];
73 edge e(u, v);
74 if(!pre[v]){
75 E.push(e);
76 int lowv = dfs(v, u);
77 lowu = min(lowv, lowu);
78 if(lowv > pre[u])
79 ans1++;
80 if(lowv >= pre[u]){
81 dcnt++;
82 dcc[dcnt].clear();
83 while(1){
84 edge x = E.top(); E.pop();
85 if(dccno[x.u] != dcnt){
86 dcc[dcnt].push_back(x.u);
87 dccno[x.u] = dcnt;
88 }
89 if(dccno[x.v] != dcnt){
90 dcc[dcnt].push_back(x.v);
91 dccno[x.v] = dcnt;
92 }
93 if(x.u == u && x.v == v) break;
94 }
95 calc(dcc[dcnt]);
96 }
97 }
98 else if(pre[v] < pre[u] && v != p)
99 lowu = min(lowu, pre[v]);
100 }
101 return lowu;
102 }
103 int main(){
104 int n, m;
105 while(scanf("%d%d", &n, &m) != EOF){
106 if(!n && !m) break;
107 init();
108 for(int i = 0; i < m; ++i){
109 int u, v;
110 scanf("%d%d", &u, &v);
111 add(u, v), add(v, u);
112 }
113 memset(dccno, 0, sizeof dccno);
114 memset(pre, 0, sizeof pre);
115 dc = dcnt = ans1 = ans2 = 0;
116 for(int i = 0; i < n; ++i){
117 if(!pre[i])
118 dfs(i, -1);
119 }
120 printf("%d %d\n", ans1, ans2);
121 }
122 return 0;
123 }
poj 1523 SPF
找割顶。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<vector>
9 #include<stack>
10 #include<queue>
11 #include<bitset>
12 using namespace std;
13
14
15 #pragma comment(linker, "/STACK:1024000000,1024000000")
16 #define inf 0x3f3f3f3f
17 #define eps 1e-12
18 #define FOR(i,s,t) for(int i = s; i < t; ++i )
19 #define REP(i,s,t) for( int i = s; i <= t; ++i )
20 #define pii pair<int,int>
21 #define MP make_pair
22 #define ls i << 1
23 #define rs ls | 1
24 #define md ((ll + rr) >> 1)
25 #define lson ll, md, ls
26 #define rson md + 1, rr, rs
27 #define LL long long
28 #define N 1020
29 #define M 2000020
30 #define Pi acos(-1.0)
31 #define mod 1000000007
32
33 int fst[N], nxt[M], vv[M], e;
34 void init(){
35 memset(fst, -1, sizeof fst); e = 0;
36 }
37 void add(int u, int v){
38 vv[e] = v, nxt[e] = fst[u], fst[u] = e++;
39 }
40
41 struct edge{
42 int u, v;
43 edge(int u = 0, int v = 0) : u(u), v(v) {}
44 };
45 stack<edge> E;
46 vector<int> dcc[N];
47 int pre[N], dccno[N], dc, dcnt;
48 bool cut[N];
49 int dfs(int u, int p){
50 int lowu = pre[u] = ++dc;
51 int son = 0;
52 for(int i = fst[u]; ~i; i = nxt[i]){
53 int v = vv[i];
54 edge e(u, v);
55 if(!pre[v]){
56 E.push(e);
57 son++;
58 int lowv = dfs(v, u);
59 lowu = min(lowu, lowv);
60 if(lowv >= pre[u]){
61 cut[u] = 1;
62 ++dcnt;
63 dcc[dcnt].clear();
64 while(1){
65 edge x = E.top(); E.pop();
66 if(dccno[x.u] != dcnt){
67 dcc[dcnt].push_back(x.u);
68 dccno[x.u] = dcnt;
69 }
70 if(dccno[x.v] != dcnt){
71 dcc[dcnt].push_back(x.v);
72 dccno[x.v] = dcnt;
73 }
74 if(x.u == u && x.v == v) break;
75 }
76 }
77 }
78 else if(pre[v] < pre[u] && v != p)
79 lowu = min(lowu, pre[v]);
80 }
81 if(p < 0 && son == 1) cut[u] = 0;
82 return lowu;
83 }
84 int num[N];
85 int main(){
86 int u, v, kk = 0;
87 while(scanf("%d", &u) != EOF){
88 if(u == 0) break;
89 init();
90 int n = u;
91 while(1){
92 scanf("%d", &v);
93 n = max(n, v);
94 add(u, v), add(v, u);
95 scanf("%d", &u);
96 n = max(n, u);
97 if(u == 0) break;
98 }
99 memset(pre, 0, sizeof pre);
100 memset(dccno, 0, sizeof dccno);
101 memset(cut, 0, sizeof cut);
102 dc = dcnt = 0;
103 for(int i = 1; i <= n; ++i)
104 if(!pre[i])
105 dfs(i, -1);
106 // printf("dcnt %d\n", dcnt);
107 memset(num, 0, sizeof num);
108 for(int i = 1; i <= dcnt; ++i){
109 for(int j = 0; j < dcc[i].size(); ++j)
110 if(cut[dcc[i][j]])
111 num[dcc[i][j]]++;
112 }
113 printf("Network #%d\n", ++kk);
114 bool ok = 0;
115 for(int i = 1; i <= n; ++i){
116 if(cut[i]){
117 printf(" SPF node %d leaves %d subnets\n", i, num[i]);
118 ok = 1;
119 }
120 }
121 if(!ok) puts(" No SPF nodes");
122 puts("");
123 }
124 }
给一个无向图。若删除某些路径，则1到n的最短路会增加，则这些路径是important的。问哪些路径是important的。
做法：对起点和终点分别求最短路，得到一定在最短路的边，然后双连通求桥。
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<cmath>
8 #include<stack>
9 #include<vector>
10 #include<queue>
11 #include<bitset>
12
13 using namespace std;
14
15
16 #pragma comment(linker, "/STACK:1024000000,1024000000")
17 #define inf 0x3f3f3f3f
18 #define eps 1e-9
19 #define FOR(i,s,t) for(int i = s; i < t; ++i )
20 #define REP(i,s,t) for( int i = s; i <= t; ++i )
21 #define pii pair<int,int>
22 #define MP make_pair
23 #define ls i << 1
24 #define rs ls | 1
25 #define md ((ll + rr) >> 1)
26 #define lson ll, md, ls
27 #define rson md + 1, rr, rs
28 #define LL long long
29 #define N 20010
30 #define M 200020
31 #define Pi acos(-1.0)
32 #define mod 1000007
33 #define ULL unsigned long long
34
35 int fst[N], vv[M], nxt[M], cost[M], e;
36 void init(){
37 memset(fst, -1, sizeof fst); e = 0;
38 }
39 void add(int u, int v, int c){
40 vv[e] = v, nxt[e] = fst[u], cost[e] = c, fst[u] = e++;
41 }
42
43 struct edge{
44 int v, id;
45 edge(int v = 0, int id = 0) : v(v), id(id) {}
46 };
47
48 vector<edge> g[N];
49 bool vis[N];
50 int d[2][N];
51
52 struct node{
53 int u, dis;
54 node(int u = 0, int dis = 0) : u(u), dis(dis) {}
55 bool operator < (const node &b) const{
56 return dis > b.dis;
57 }
58 };
59 void dij(int s, int k){
60 memset(vis, 0, sizeof vis);
61 for(int i = 0; i < N; ++i)
62 d[k][i] = (1LL<<31) - 1;
63 priority_queue<node> q;
64 q.push(node(s, 0));
65 d[k][s] = 0;
66 while(!q.empty()){
67 node p = q.top(); q.pop();
68 int u = p.u, dis = p.dis;
69 if(vis[u]) continue;
70 vis[u] = 1;
71 for(int i = fst[u]; ~i; i = nxt[i]){
72 int v = vv[i], c = cost[i];
73 if(d[k][v] > d[k][u] + c){
74 d[k][v] = d[k][u] + c;
75 q.push(node(v, d[k][v]));
76 }
77 }
78 }
79 }
80 int fa[N], id[N];
81 void bfs(int s){
82 queue<int> q;
83 q.push(s);
84 memset(fa, -1, sizeof fa);
85 memset(vis, 0, sizeof vis);
86 vis[s] = 1;
87 while(!q.empty()){
88 int u = q.front(); q.pop();
89 for(int i = 0; i < g[u].size(); ++i){
90 edge e = g[u][i];
91 int v = e.v;
92 if(!vis[v]){
93 fa[v] = u, id[v] = e.id;
94 vis[v] = 1;
95 q.push(v);
96 }
97 }
98 }
99 }
100 int pre[N], dc, cut[M];
101 int dfs(int u, int cot){
102 int lowu = pre[u] = ++dc;
103 for(int i = 0; i < g[u].size(); ++i){
104 edge e = g[u][i];
105 int v = e.v;
106 if(!pre[v]){
107 int lowv = dfs(v, e.id);
108 lowu = min(lowu, lowv);
109 if(lowv > pre[u])
110 cut[e.id] = 1;
111 }
112 else if(cot != e.id)
113 lowu = min(lowu, pre[v]);
114 }
115 return lowu;
116 }
117
118 int main(){
119 // freopen("tt.txt", "r", stdin);
120 freopen("important.in", "r", stdin);
121 freopen("important.out", "w", stdout);
122 int n, m;
123 while(scanf("%d%d", &n, &m) != EOF){
124 init();
125 for(int i = 0; i < m; ++i){
126 int u, v, c;
127 scanf("%d%d%d", &u, &v, &c);
128 add(u, v, c), add(v, u, c);
129 }
130 dij(1, 0);
131 dij(n, 1);
132 int tot = d[0][n];
133 for(int i = 1; i <= n; ++i)
134 g[i].clear();
135 for(int i = 1; i <= n; ++i){
136 for(int j = fst[i]; ~j; j = nxt[j]){
137 int u = i, v = vv[j], c = cost[j];
138 if(d[0][u] + c + d[1][v] == tot || d[0][v] + c + d[1][u] == tot){
139 g[u].push_back(edge(v, j / 2));
140 g[v].push_back(edge(u, j / 2));
141 }
142 }
143 }
144 bfs(1);
145 dc = 0;
146 memset(cut, 0, sizeof cut);
147 memset(pre, 0, sizeof pre);
148 dfs(1, -1);
149 int u = n;
150 vector<int> ans;
151 while(u != 1){
152 if(cut[id[u]]) ans.push_back(id[u]+1);
153 u = fa[u];
154 }
155 printf("%d\n", ans.size());
156 if(ans.size() == 0){
157 puts(""); continue;
158 }
159 sort(ans.begin(), ans.end());
160 for(int i = 0; i < ans.size(); ++i)
161 printf("%d%c", ans[i], i == ans.size() - 1 ? ‘\n‘ : ‘ ‘);
162 }
163 return 0;
164 }
时间： 2024-11-03 22:23:53