题目大意：
给出一个带权有向图，将其中的强连通分量缩成点，再求最短路。若有输出最小值，若没有输出：“Nao e possivel entregar a carta” (“It’s impossible to deliver the letter”).
解题思路：
1、建图，然后Tarjan算法缩点。
2、求最短路然后输出。
下面是代码：
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int MAXN = 505;
const int inf = 1<<30;
struct node
{
int v,w,next;
} edge[MAXN*MAXN],nedge[MAXN*MAXN];
int dfn[MAXN],low[MAXN],vis[MAXN],stack1[MAXN],head[MAXN],top,time,cnt;
int num[MAXN],numcnt;
int nhead[MAXN],ncnt;
int dis[MAXN];
int n,m;
void init()
{
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
memset(head,-1,sizeof(head));
memset(nhead,-1,sizeof(nhead));
memset(stack1,0,sizeof(stack1));
ncnt=0;
top=0;
cnt=0;
time=1;
numcnt=1;
}
void addedge(int u,int v,int w)
{
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt;
cnt++;
}
int min(int a , int b)
{
if(a>b)a=b;
return a;
}
void dfs(int u,int fa)
{
dfn[u]=time;
low[u]=time;
time++;
vis[u]=1;
stack1[top]=u;
top++;
for(int i = head[u] ; i != -1 ; i = edge[i].next)
{
int v=edge[i].v;
if(!vis[v])
{
dfs(v,u);
low[u]=min(low[u],low[v]);
}
else if(vis[v]==1)
{
low[u]=min(low[u],dfn[v]);
}
}
if(dfn[u]==low[u])
{
while(stack1[top]!=u&&top>0)
{
top--;
num[stack1[top]]=numcnt;
vis[stack1[top]]=2;
}
numcnt++;
}
}
void Tarjan()
{
for(int i = 1 ; i<=n; i++)
{
if(!vis[i])dfs(i,0);
}
}
void Build()
{
int u,v,w;
for(int i = 0 ; i < m ; i ++ )
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
}
}
void Readdedge(int u ,int v ,int w)
{
nedge[ncnt].v=v;
nedge[ncnt].w=w;
nedge[ncnt].next=nhead[u];
nhead[u]=ncnt;
ncnt++;
}
void ReBuild()
{
for(int i = 1 ; i <= n ; i++ )
{
for(int j = head[i] ; j != -1 ; j=edge[j].next )
{
int v=edge[j].v;
if(num[i]!=num[v])
{
Readdedge(num[i],num[v],edge[j].w);
}
}
}
}
void spfa(int start,int end)
{
queue <int > q;
for(int i =1; i<numcnt; i++)
{
dis[i]=inf;
vis[i]=0;
}
dis[start]=0;
vis[start]=1;
q.push(start);
while(!q.empty())
{
int t = q.front();
int p = nhead[t];
vis[t]=0;
q.pop();
while(p!=-1)
{
int v = nedge[p].v;
int w = nedge[p].w;
if(dis[v]>dis[t]+w)
{
dis[v]=dis[t]+w;
if(!vis[v])
{
q.push(v);
vis[v]=1;
}
}
p=nedge[p].next;
}
}
if(dis[end]!=inf)
{
printf("%d\n",dis[end]);
}
else
{
printf("Nao e possivel entregar a carta\n");
}
}
void Ask()
{
int k,u,v;
scanf("%d",&k);
while(k--)
{
scanf("%d%d",&u,&v);
if(num[u]==num[v])
{
puts("0");
continue;
}
spfa(num[u],num[v]);
}
}
int main()
{
while(scanf("%d%d",&n,&m),n)
{
init();
Build();
Tarjan();
ReBuild();
Ask();
puts("");
}
return 0;
}