题目大意：给出N个点，M条有向边，问是否任意两点u,v都满足u能到达v或者v能到达u
解题思路：强连通分量内的所有的点都满足，接着要判断一下其他的点能否满足了
求出所有的强连通分量，接着缩点，用桥连接，形成新的图(以下所说的点都是指新的图的点)
如果一个点同时指向另外两个不同的点，那么这两个点之间肯定是不能相互到达的，所以拓扑排序一下，就可以知道是否符合了
#include
#include
#define min(a,b) ((a) < (b) ? (a) : (b)) #define N 10100 #define M 60100 struct Edge{ int from, to, next; }E[M]; struct Node{ int x, y; }node[M]; int n, m, tot, dfs_clock, top, scc_cnt; int head[N], pre[N], stack[N], sccno[N], lowlink[N], in[N]; void AddEdge(int u, int v) { E[tot].from = u; E[tot].to = v; E[tot].next = head[u]; head[u] = tot++; } void init() { scanf(%d%d, &n, &m); memset(head, -1, sizeof(head)); tot = 0; int u, v; for (int i = 0; i < m; i++) { scanf(%d%d, &node[i].x, &node[i].y); AddEdge(node[i].x, node[i].y); } } void dfs(int u) { pre[u] = lowlink[u] = ++dfs_clock; stack[++top] = u; int v; for (int i = head[u]; i != -1; i = E[i].next) { v = E[i].to; if (!pre[v]) { dfs(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) { lowlink[u] = min(lowlink[u], pre[v]); } } if (pre[u] == lowlink[u]) { scc_cnt++; while (1) { v = stack[top--]; sccno[v] = scc_cnt; if (v == u) break; } } } int find() { int cnt = 0, num; for (int i = 1; i <= scc_cnt; i++) { if (in[i] == 0) { cnt++; num = i; } } if (cnt == 1) return num; return 0; } bool TopoOrder(){ int cnt = 0; int u; while (u = find()) { in[u] = -1; for (int i = head[u]; i != -1; i = E[i].next) { in[E[i].to]--; } cnt++; } return cnt == scc_cnt; } void solve() { memset(pre, 0, sizeof(pre)); memset(sccno, 0, sizeof(sccno)); dfs_clock = top = scc_cnt = 0; for (int i = 1; i <= n; i++) if (!pre[i]) dfs(i); if (scc_cnt == 1) { printf(Yes ); return ; } memset(in, 0, sizeof(in)); memset(head, -1, sizeof(head)); tot = 0; int u, v; for (int i = 0; i < m; i++) { u = sccno[node[i].x]; v = sccno[node[i].y]; if (u != v) { in[v]++; AddEdge(u, v); } } int ans = 0; for (int i = 1; i <= scc_cnt; i++) if (in[i] == 0) ans++; if (ans > 1) { printf(No ); return ; } if (TopoOrder()) printf(Yes ); else printf(No ); } int main() { int test; scanf(%d, &test); while (test--) { init(); solve(); } return 0; }
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