poj3255 Roadblocks--次短路spfa

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原题链接：http://poj.org/problem?id=3255

题意：n个点，标号为1到n，m条路，u，v，len，表示u与v之间路长为len，求1到n第二短路长，题目保证存在第二短路径。

#define _CRT_SECURE_NO_DEPRECATE #include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#define INF 99999999#define eps 0.0001using namespace std;struct Edge{int v;int len;int next;};int n, m;int cnt;Edge edge[200010];int head[5010];bool vis[5010];int dis[5010];int rdis[5010];void add(int u, int v, int len){edge[cnt].v = v; edge[cnt].len = len; edge[cnt].next = head[u]; head[u] = cnt++;edge[cnt].v = u; edge[cnt].len = len; edge[cnt].next = head[v]; head[v] = cnt++;}void spfa(int s, int d[]){memset(vis, 0, sizeof(vis));queue<int> Q;d[s] = 0;vis[s] = 1;Q.push(s);while (!Q.empty()){int u = Q.front();vis[u] = 0;Q.pop();for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if (d[v] > d[u] + edge[i].len){d[v] = d[u] + edge[i].len;if (!vis[v]){vis[v] = 1;Q.push(v);}}}}}int main(){int u, v, len;while (~scanf("%d%d", &n, &m)){//initmemset(head, -1, sizeof(head));cnt = 0;fill(dis, dis + n + 1, INF);fill(rdis, rdis + n + 1, INF);while (m--){scanf("%d%d%d", &u, &v, &len);add(u, v, len);}spfa(1, dis);spfa(n, rdis);int ans = INF;for (int i = 1; i <= n; i++){for (int j = head[i]; j != -1; j = edge[j].next){int v = edge[j].v;int len = edge[j].len;int d = dis[i] + rdis[v] + len;if (d > dis[n] && d < ans)ans = d;}}printf("%d\n", ans);}return 0;}

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