题意：给出n个点，m条边权为1的无向边，破坏最多的道路，使得从s1到t1,s2到t2的距离不超过d1,d2
因为最后s1,t1是连通的，且要破坏掉最多的道路，那么就是求s1到t1之间的最短路
用bfs求出任意两个顶点之间的距离， 如果d[s1][t1]>d1||d[s2][t2]>d2，那么不可能
然后就枚举重叠的区间（就像题解里面说的"H"形一样）
如果枚举区间是1--n,1--i的话，需要交换四次
如果枚举区间是1--n,1--n的话，则只需要交换一次就可以了
看的这一篇题解：http://www.cnblogs.com/qscqesze/p/4487498.html
交换一次的
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include <cmath>
5 #include<stack>
6 #include<vector>
7 #include<map>
8 #include<set>
9 #include<queue>
10 #include<algorithm>
11 using namespace std;
12
13 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
14
15 typedef long long LL;
16 const int INF = (1<<30)-1;
17 const int mod=1000000007;
18 const int maxn=5005;
19
20 int n,m;
21 vector<int> e[maxn];
22 int vis[maxn];
23 int d[maxn][maxn];
24
25
26 int main(){
27 scanf("%d %d",&n,&m);
28 for(int i=1;i<=m;i++){
29 int u,v;
30 scanf("%d %d",&u,&v);
31 e[u].push_back(v);
32 e[v].push_back(u);
33 }
34
35 int s1,t1,d1,s2,t2,d2;
36 scanf("%d %d %d %d %d %d",&s1,&t1,&d1,&s2,&t2,&d2);
37
38
39 for(int i=1;i<=n;i++){
40 queue<int> q;
41 memset(vis,0,sizeof(vis));
42 vis[i]=1;
43 q.push(i);
44
45 while(!q.empty()){
46 int u=q.front();q.pop();
47
48 for(int j=0;j<e[u].size();j++){
49 int v=e[u][j];
50 if(vis[v]) continue;
51 vis[v]=1;
52 d[i][v]=d[i][u]+1;
53 q.push(v);
54 }
55 }
56 }
57
58 if(d[s1][t1]>d1||d[s2][t2]>d2) {
59 puts("-1");
60 return 0;
61 }
62
63
64 int ans=d[s1][t1]+d[s2][t2];
65 for(int i=1;i<=n;i++){
66 for(int j=1;j<=n;j++){
67 if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
68 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);
69
70 if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
71 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);
72 }
73 }
74 printf("%d\n",m-ans);
75 return 0;
76 }
交换四次的
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include <cmath>
5 #include<stack>
6 #include<vector>
7 #include<map>
8 #include<set>
9 #include<queue>
10 #include<algorithm>
11 using namespace std;
12
13 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
14
15 typedef long long LL;
16 const int INF = (1<<30)-1;
17 const int mod=1000000007;
18 const int maxn=5005;
19
20 int n,m;
21 vector<int> e[maxn];
22 int vis[maxn];
23 int d[maxn][maxn];
24
25
26 int main(){
27 scanf("%d %d",&n,&m);
28 for(int i=1;i<=m;i++){
29 int u,v;
30 scanf("%d %d",&u,&v);
31 e[u].push_back(v);
32 e[v].push_back(u);
33 }
34
35 int s1,t1,d1,s2,t2,d2;
36 scanf("%d %d %d %d %d %d",&s1,&t1,&d1,&s2,&t2,&d2);
37
38
39 for(int i=1;i<=n;i++){
40 queue<int> q;
41 memset(vis,0,sizeof(vis));
42 vis[i]=1;
43 q.push(i);
44
45 while(!q.empty()){
46 int u=q.front();q.pop();
47
48 for(int j=0;j<e[u].size();j++){
49 int v=e[u][j];
50 if(vis[v]) continue;
51 vis[v]=1;
52 d[i][v]=d[i][u]+1;
53 q.push(v);
54 }
55 }
56 }
57
58 if(d[s1][t1]>d1||d[s2][t2]>d2) {
59 puts("-1");
60 return 0;
61 }
62
63
64 int ans=d[s1][t1]+d[s2][t2];
65 for(int i=1;i<=n;i++){
66 for(int j=1;j<=i;j++){
67 if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
68 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);
69
70 if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
71 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);
72
73 if(d[t1][i]+d[i][j]+d[j][s1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
74 ans=min(ans,d[t1][i]+d[i][j]+d[j][s1]+d[s2][i]+d[j][t2]);
75
76 if(d[t1][i]+d[i][j]+d[j][s1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
77 ans=min(ans,d[t1][i]+d[i][j]+d[j][s1]+d[t2][i]+d[j][s2]);
78 }
79 }
80 printf("%d\n",m-ans);
81 return 0;
82 }
加油---g00000000----
时间： 2024-10-10 13:43:54