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
#include
const 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");
}
}