53 lines
1.3 KiB
C++
53 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 200010, M = N << 1;
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int n;
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bool st[N];
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int ans = INF;
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int root, ed;
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// Tarjan算法求割点
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int dfn[N], low[N], ts;
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void tarjan(int u, int fa) {
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low[u] = dfn[u] = ++ts;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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if (!dfn[v]) {
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tarjan(v, u);
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low[u] = min(low[u], low[v]);
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// u是割点
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if (u != root && low[v] >= dfn[u]) {
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if (u == ed) continue; // 根据题意，u必须在root和ed之间
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if (dfn[ed] >= dfn[v]) ans = min(ans, u);
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}
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} else
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low[u] = min(low[u], dfn[v]);
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}
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}
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int main() {
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// 初始化链式前向星
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memset(h, -1, sizeof h);
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scanf("%d", &n);
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int x, y;
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while (scanf("%d %d", &x, &y), x || y)
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if (x != y) add(x, y), add(y, x);
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scanf("%d %d", &root, &ed);
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tarjan(root, -1); // 从其中一个信息中心开始遍历
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if (ans == INF)
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printf("No solution");
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else
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printf("%d", ans);
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return 0;
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} |