poj2553--Tarjan

poj2553--Tarjan算法

题目大意：

    此题主要难点在于题目的理解：A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w.

sink 就是指某个能到达的点也能到达他自己的，即他们在同一个强联通分量中。

题目解法：

    找到出度为0的强联通分量中的所有点都是符合要求的。题目还要求将最终答案排序后输出，这里wa了一次，我以为是每个强联通分量单独排序。

题目源码：

//#define LOCAL#include <stdio.h>#include <string.h>#include <queue>using namespace std;#define MAXN 5500#define INF 0x3f3f3f3fint n, m;struct Edge{int to;int next;};Edge edge[MAXN * MAXN];int head[MAXN], e;int dfn[MAXN], low[MAXN], cnt;int stack[MAXN], top;int instack[MAXN];int belong[MAXN], index;int out[MAXN];int ans[MAXN];int min(int a, int b){return a < b ? a : b;}void addEdge(int u, int v){edge[e].to = v;edge[e].next = head[u];head[u] = e++;}void init(){int i, j;int u, v;e = 0;cnt = 1;top = 0;index = 0;memset(head, -1, sizeof(head)); memset(dfn, -1, sizeof(dfn));memset(low, -1, sizeof(low)); memset(instack, 0, sizeof(instack));memset(belong, 0, sizeof(belong));memset(out, 0, sizeof(out)); memset(ans, 0, sizeof(ans));for(i = 1; i <= m; i++){scanf("%d%d", &u, &v);addEdge(u, v);} }void tarjan(int u){int i, v;low[u] = dfn[u] = cnt++;instack[u] = 1;stack[++top] = u;for(i = head[u]; i != -1; i = edge[i].next){v = edge[i].to;if(dfn[v] == -1){tarjan(v);low[u] = min(low[u], low[v]);} else if(instack[v])low[u] = min(low[u], dfn[v]);}if(low[u] == dfn[u]){index++;do{v = stack[top--];belong[v] = index;}while(top != 0 && v != u);}}void solve(){int i, j;int v;int flag;while(scanf("%d%d", &n, &m) && n){flag = 1;init(); // 获得head数组及edge数组 for(i = 1; i <= n; i++)if(dfn[i] == -1)tarjan(i); // 获得belong数组 for(i = 1; i <= n; i++){for(j = head[i]; j != -1; j = edge[j].next) // 获得out数组 {v = edge[j].to;if(belong[i] != belong[v])out[belong[i]]++;}}flag = 1;for(i = 1; i <= n; i++){if(!out[belong[i]]){if(flag){printf("%d", i);flag = 0;}else{printf(" %d", i);}}}printf("\n");}}int main(){#ifdef LOCALfreopen("poj2553.txt", "r", stdin);freopen("poj2553Out.txt", "w", stdout);#endifsolve();return 0;}