POJ 3114 Countries in War

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;
}