poj 3114 Countries in War（Tarjan+floyd）

题目大意是这样说的，在一个有向图中，每两点间通信需要一定的时间，但同一个强连通分量里传递信息不用时间，给两点u，v求他们最小的通信时间。

思路：题意比较直白，就是缩点后，重新建图，求其最短路。看了一下点的范围是500，查询次数是100，想了一下，用SPFA求最短路时间会快点，但floyd也可以，代码短。最后发现时间有点多。。。。

//2644K    985MS#include#includeconst int VM = 505;const int EM = VM*VM;const int inf = 0x3f3f3f3f;struct Edge{    int to,w,nxt;}edge[EM];int head[VM],vis[VM],stack[VM],belong[VM];int dfn[VM],low[VM],mat[VM][VM];int ep,cnt,scc,top,n;int min(int a,int b){    return a > b ? b : a;}void addedge (int cu,int cv,int cw){    edge[ep].to = cv;    edge[ep].w = cw;    edge[ep].nxt = head[cu];    head[cu] = ep ++;}void Tarjan (int u){    int v;    dfn[u] = low[u] = ++cnt;    vis[u] = 1;    stack[top ++] = u;    for (int i = head[u];i != -1;i = edge[i].nxt)    {        v = edge[i].to;        if (!dfn[v])        {            Tarjan(v);            low[u] = min(low[u],low[v]);        }        else if (vis[v]) low[u] = min(low[u],dfn[v]);    }    if (dfn[u] == low[u])    {        ++scc;        do{            v = stack[--top];            vis[v] = 0;            belong[v] = scc;        }while (u != v);    }}void solve(){    memset (dfn,0,sizeof(dfn));    memset (vis,0,sizeof(vis));    memset (mat,0x3f,sizeof(mat));    scc = top = cnt = 0;    int u,v;    for (u = 1;u <= n;u ++)        if (!dfn[u])            Tarjan(u);    for (u = 1;u <= n;u ++)    {        for (int i = head[u];i != -1;i = edge[i].nxt)        {            v = edge[i].to;            if (belong[u] != belong[v])                mat[belong[u]][belong[v]] = min(mat[belong[u]][belong[v]],edge[i].w);        }    }    for (u = 1;u <= scc;u ++) mat[u][u] = 0;}void floyd(){    for (int k = 1;k <= scc;k ++)        for (int i = 1;i <= scc;i ++)            for (int j = 1;j <= scc;j ++)                if (mat[i][j] > mat[i][k] + mat[k][j])                    mat[i][j] = mat[i][k] + mat[k][j];}int main (){    #ifdef LOCAL        freopen ("in.txt","r",stdin);    #endif    int m,u,v,w;    while (~scanf ("%d%d",&n,&m)&&n)    {        memset (head,-1,sizeof(head));        ep = 0;        while (m --)        {            scanf ("%d%d%d",&u,&v,&w);            addedge (u,v,w);        }        solve ();        floyd();        scanf ("%d",&m);        while (m --)        {            scanf ("%d %d",&u,&v);            u = belong[u],v = belong[v];            if (mat[u][v]!= inf)                printf ("%d\n",mat[u][v]);            else printf ("Nao e possivel entregar a carta\n");        }        printf ("\n");    }}