这题做了几个小时，基本思路肯定是求两点路径中的割点数目，思路是tarjan缩点，然后以割点和连通块作为新节点见图。转化为lca求解。
结合点——双连通分量与LCA。
1 /* 3686 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <algorithm>
12 #include <cstdio>
13 #include <cmath>
14 #include <ctime>
15 #include <cstring>
16 #include <climits>
17 #include <cctype>
18 #include <cassert>
19 #include <functional>
20 #include <iterator>
21 #include <iomanip>
22 using namespace std;
23 //#pragma comment(linker,"/STACK:102400000,1024000")
24
25 #define sti set<int>
26 #define stpii set<pair<int, int> >
27 #define mpii map<int,int>
28 #define vi vector<int>
29 #define pii pair<int,int>
30 #define vpii vector<pair<int,int> >
31 #define rep(i, a, n) for (int i=a;i<n;++i)
32 #define per(i, a, n) for (int i=n-1;i>=a;--i)
33 #define clr clear
34 #define pb push_back
35 #define mp make_pair
36 #define fir first
37 #define sec second
38 #define all(x) (x).begin(),(x).end()
39 #define SZ(x) ((int)(x).size())
40 #define lson l, mid, rt<<1
41 #define rson mid+1, r, rt<<1|1
42
43 typedef struct {
44 int u, v, f, nxt;
45 } edge_t;
46
47 typedef struct {
48 int v, nxt;
49 } edge;
50
51 const int maxv = 10005;
52 const int maxe = 200005;
53 int head[maxv], l, top;
54 int pre[maxv], low[maxv];
55 bool iscut[maxv];
56 int cnt[maxv], dfs_clock, bcc_cnt;
57 int bn[maxe];
58 int S[maxe];
59 vi bcc[maxv];
60 edge_t E[maxe];
61 int n, m;
62
63 const int maxvv = maxv * 2;
64 int mark[maxvv];
65 int head_[maxvv], l_;
66 int cutn[maxvv];
67 edge E_[maxe];
68
69 int deep[maxvv], beg[maxvv];
70 int V[maxvv<<1], D[maxvv<<1];
71
72 int dp[16][maxvv<<1];
73
74 void init_() {
75 memset(head_, -1, sizeof(head_));
76 memset(mark, 0, sizeof(mark));
77 l_ = 0;
78 }
79
80 void addEdge_(int u, int v) {
81 E_[l_].v = v;
82 E_[l_].nxt = head_[u];
83 head_[u] = l_++;
84
85 E_[l_].v = u;
86 E_[l_].nxt = head_[v];
87 head_[v] = l_++;
88 }
89
90 void init() {
91 l = dfs_clock = bcc_cnt = top = 0;
92 memset(head, -1, sizeof(head));
93 memset(iscut, false, sizeof(iscut));
94 memset(cnt, 0, sizeof(cnt));
95 memset(cutn, 0, sizeof(cutn));
96 memset(pre, 0, sizeof(pre));
97 rep(i, 1, n+1)
98 bcc[i].clr();
99 }
100
101 void addEdge(int u, int v) {
102 E[l].u = u;
103 E[l].f = 0;
104 E[l].v = v;
105 E[l].nxt = head[u];
106 head[u] = l++;
107
108 E[l].u = v;
109 E[l].f = 0;
110 E[l].v = u;
111 E[l].nxt = head[v];
112 head[v] = l++;
113 }
114
115 void tarjan(int u, int fa) {
116 int v, k;
117
118 low[u] = pre[u] = ++dfs_clock;
119 for (k=head[u]; k!=-1; k=E[k].nxt) {
120 if (E[k].f)
121 continue;
122 E[k].f = E[k^1].f = 1;
123 v = E[k].v;
124 S[top++] = k;
125 if (!pre[v]) {
126 tarjan(v, u);
127 low[u] = min(low[u], low[v]);
128 if (low[v] >= pre[u]) {
129 iscut[u] = true;
130 ++cnt[u];
131 bcc_cnt++;
132 while (1) {
133 int kk = S[--top];
134 bn[kk>>1] = bcc_cnt;
135 bcc[E[kk].u].pb(bcc_cnt);
136 bcc[E[kk].v].pb(bcc_cnt);
137 if (kk == k)
138 break;
139 }
140 }
141 } else {
142 low[u] = min(low[u], pre[v]);
143 }
144 }
145 }
146
147 void dfs(int u, int fa, int d) {
148 mark[u] = dfs_clock;
149 deep[u] = d;
150 V[++top] = u;
151 D[top] = d;
152 beg[u] = top;
153
154 int v, k;
155
156 for (k=head_[u]; k!=-1; k=E_[k].nxt) {
157 v = E_[k].v;
158 if (v == fa)
159 continue;
160 dfs(v, u, d+1);
161 V[++top] = u;
162 D[top] = d;
163 }
164 }
165
166 void init_RMQ(int n) {
167 int i, j;
168
169 for (i=1; i<=n; ++i)
170 dp[0][i] = i;
171 for (j=1; (1<<j)<=n; ++j)
172 for (i=1; i+(1<<j)-1<=n; ++i)
173 if (D[dp[j-1][i]] < D[dp[j-1][i+(1<<(j-1))]])
174 dp[j][i] = dp[j-1][i];
175 else
176 dp[j][i] = dp[j-1][i+(1<<(j-1))];
177 }
178
179 int RMQ(int l, int r) {
180 if (l > r)
181 swap(l, r);
182
183 int k = 0;
184
185 while (1<<(k+1) <= r-l+1)
186 ++k;
187
188 if (D[dp[k][l]] < D[dp[k][r-(1<<k)+1]])
189 return V[dp[k][l]];
190 else
191 return V[dp[k][r-(1<<k)+1]];
192 }
193
194 void solve() {
195 int u, v, lca;
196
197 rep(i, 1, n+1) {
198 if (!pre[i]) {
199 tarjan(i, -1);
200 if (cnt[i] <= 1)
201 iscut[i] = false;
202 }
203 }
204
205 int cid = bcc_cnt;
206
207 init_();
208 cid = bcc_cnt;
209 for (u=1; u<=n; ++u) {
210 if (!iscut[u])
211 continue;
212 sort(all(bcc[u]));
213 cutn[++cid] = 1;
214 addEdge_(cid, bcc[u][0]);
215 int sz = SZ(bcc[u]);
216 rep(i, 1, sz) {
217 if (bcc[u][i] != bcc[u][i-1]) {
218 addEdge_(cid, bcc[u][i]);
219 }
220 }
221 }
222
223 top = 0;
224 ++dfs_clock;
225 rep(i, 1, cid+1) {
226 if (mark[i] != dfs_clock)
227 dfs(i, 0, 0);
228 }
229
230 init_RMQ(top);
231
232 int q;
233 int ans;
234
235 scanf("%d", &q);
236 while (q--) {
237 scanf("%d %d", &u, &v);
238 u = bn[u-1];
239 v = bn[v-1];
240 lca = RMQ(beg[u], beg[v]);
241 ans = (deep[u]+deep[v] - deep[lca]*2 + 1) / 2;
242 printf("%d\n", ans);
243 }
244 }
245
246 int main() {
247 ios::sync_with_stdio(false);
248 #ifndef ONLINE_JUDGE
249 freopen("data.in", "r", stdin);
250 freopen("data.out", "w", stdout);
251 #endif
252
253 int u, v;
254
255 while (scanf("%d %d", &n, &m)!=EOF && (n||m)) {
256 init();
257 rep(i, 0, m) {
258 scanf("%d %d", &u, &v);
259 addEdge(u, v);
260 }
261
262 solve();
263 }
264
265 #ifndef ONLINE_JUDGE
266 printf("time = %d.\n", (int)clock());
267 #endif
268
269 return 0;
270 }
时间： 2024-10-25 14:55:59