SPFA POJ 3268 Silver Cow Party

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题目链接：http://poj.org/problem?id=3268

题目大意：有n个位置，m条路，求所有点到某一固定点x往返时间的最大值（有向），从i位置到j位置所需要用的时间

算法：SPFA

思路：2次SPFA，一次正向，第二次反向

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<queue>using namespace std;const int inf = 0xfffffff;queue<int> q;int head[2][123456];struct Edge{int v, w, next;}edge[2][168971];int dis[123456];int vis[123456];int tot[2];int ans[123465];int n;void add_edge(int s, int v, int w, int k){edge[k][tot[k]].v = v;edge[k][tot[k]].w = w;edge[k][tot[k]].next = head[k][s];head[k][s] = tot[k] ++;}void SPFA(int pos, int k){int i;for(i = 1; i <= n; i ++)dis[i] = inf ;q.push(pos);vis[pos] = 1;dis[pos] = 0;while(!q.empty()){int u = q.front();q.pop();vis[u] = 0;for(i = head[k][u]; i != -1; i = edge[k][i].next){int e = edge[k][i].v;if(dis[e] > dis[u] + edge[k][i].w){dis[e] = dis[u] + edge[k][i].w;if(!vis[e])q.push(e);vis[e] = 1;}}}for(i = 1; i <= n; i ++){//printf("dis[%d] = %d\n", i, dis[i]);if(i != pos)ans[i] += dis[i];}}int main(){int m, k;while(scanf("%d%d%d", &n, &m, &k) != EOF){int a, b, c, i;tot[1] = tot[0] = 0;memset(head, -1, sizeof(head));memset(ans, 0, sizeof(ans));for(i = 1; i <= m; i ++){scanf("%d%d%d", &a, &b, &c);add_edge(a, b, c, 1);add_edge(b, a, c, 0);}memset(vis, 0, sizeof(vis));SPFA(k, 0);memset(vis, 0, sizeof(vis));SPFA(k, 1);int minc = -inf;for(i = 1; i <= n; i ++){if(i != k){if(minc < ans[i])minc = ans[i];}}printf("%d\n", minc);}return 0;}