89 lines
2.7 KiB
C++
89 lines
2.7 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 10010, M = 60010;
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int dnt[N], low[N], id[N], sz[N], timestamp, scc_cnt;
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int stk[N], top, in_stk[N];
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int dist[N]; //记录强连通分量距离起点的距离
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int n, m;
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int h[N], hs[N], ne[M], e[M], w[M], idx; // hs[u]为强连通分量建的图的表头
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void add(int h[], int a, int b, int c) {
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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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//有向图的Tarjan算法求强连通分量
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void tarjan(int u) {
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low[u] = dnt[u] = ++timestamp;
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stk[++top] = u;
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in_stk[u] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!dnt[j]) {
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tarjan(j);
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low[u] = min(low[u], low[j]);
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} else if (in_stk[j])
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low[u] = min(low[u], dnt[j]);
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}
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if (low[u] == dnt[u]) {
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int x;
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++scc_cnt;
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do {
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x = stk[top--];
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in_stk[x] = false;
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id[x] = scc_cnt;
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sz[scc_cnt]++;
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} while (x != u);
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}
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}
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int main() {
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memset(h, -1, sizeof h);
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memset(hs, -1, sizeof hs);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(h, b, a, 1); // b->边权为1的边->a
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}
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//超级源点0号点
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for (int i = 1; i <= n; i++) add(h, 0, i, 100); // 0号点有一条边权100的边到i
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// Tarjan算法求强连通分量(有向图)
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for (int i = 0; i <= n; i++)
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if (!dnt[i]) tarjan(i); //对于每个未访问过的点进行dfs,防止有孤立的点访问不到
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if (scc_cnt != n + 1)
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puts("Poor Xed");
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else {
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for (int u = 0; u <= n; ++u) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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int a = id[u], b = id[j];
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if (a != b)
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add(hs, a, b, w[i]); //建新图
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}
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}
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dist[0] = 0;
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//递推求出新建的图中的最长路(按照拓扑序来递推，scc_cnt ~ 1这个顺序符合拓扑序)
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// Tarjan算法本质是一个dfs算法，得到的拓扑序是倒序的，需要倒序枚举
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//拓扑序Topsort本质是一个bfs算法，得到的拓扑序是正序的，需要正序枚举
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for (int u = scc_cnt; u >= 1; u--) {
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//枚举i邻接的所有的边，找出最大的状态转移
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for (int i = hs[u]; ~i; i = ne[i]) {
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int j = e[i];
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dist[j] = max(dist[j], dist[u] + w[i]);
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}
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}
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int res = 0;
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for (int i = 1; i <= scc_cnt; i++) res += (sz[i] * dist[i]);
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printf("%d\n", res);
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}
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return 0;
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} |