poj 3114 Countries in War（缩点+最短路）

Countries in War

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1932 Accepted: 607

Description

In the year 2050, after different attempts of the UN to maintain peace in the world, the third world war broke out. The importance of industrial, commercial and military secrets obliges all the countries to use extremely sophisticated espionage services, so that each city in the world has at least one spy of each country. These spies need to communicate with other spies, informers as well as their headquarters during their actions. Unluckily there doesn’t exist a secure way for a spy to communicate during the war period, therefore the messages are always sent in code so that only the addressee is able to read the message and understand its meaning.

The spies use the only service that functions during the war period, the post. Each city has a postal agency where the letters are sent. The letters can be sent directly to their destination or to other postal agencies, until the letter arrives at the postal agency of the destination city, if possible.

The postal agency in city A can send a printed letter to the postal agency in city B if there is an agreement on sending letters, which determines the time, in hours, that a letter takes to reach city B from city A (and not necessarily the opposite). If there is no agreement between the agencies A and B, the agency A can try to send the letter to any agency so that the letter can reach its destination as early as possible

Some agencies are connected with electronic communication media, such as satellites and optical fibers. Before the war, these connections could reach all the agencies, making that a letter could be sent instantly. But during the period of hostilities every country starts to control electronic communication and an agency can only send a letter to another agency by electronic media (or instantly) if they are in the same country. Two agencies, A and B, are in the same country if a printed letter sent from any one of the agencies can be delivered to the other one.

The espionage service of your country has managed to obtain the content of all the agreements on sending messages existing in the world and desires to find out the minimum time to send a letter between different pairs of cities. Are you capable of helping them?

Input

The input contains several test cases. The first line of each test case contains two integer separated by a space,N (1 ≤ N ≤ 500) and E (0 ≤ E ≤ N²), indicating the numbers of cities (numbered from 1 toN) and of agreements on sending messages, respectively. Following them, then,E lines, each containing three integers separated by spaces, X,Y and H (1 ≤ X, Y ≤ N, 1 ≤ H ≤ 1000), indicating that there exist an agreement to send a printed letter from cityX to city Y, and that such a letter will be delivered in H hours.

After that, there will be a line with an integer K (0 ≤ K ≤ 100), the number of queries. Finally, there will beK lines, each representing a query and containing two integers separated by a space,O and D (1 ≤ O, D ≤ N). You must determine the minimum time to send a letter from cityO to city D.

The end of the input is indicated by N = 0.

Output

For each test case your program should produce K lines of output. TheI-th line should contain an integer M, the minimum time, in hours, to send a letter in theI-th query. If there aren’t communication media between the cities of the query, you should print “Nao e possivel entregar a carta” (“It’s impossible to deliver the letter”).

Print a blank line after each test case.

Sample Input

4 51 2 52 1 103 4 84 3 72 3 651 21 31 44 34 13 31 2 102 3 13 2 131 33 13 20 0

Sample Output

0660Nao e possivel entregar a carta10Nao e possivel entregar a carta0

题意：给出一个有向图，若两个点在同一强连通分量中，则这两个点的距离为0.给出若干询问u v，求u到v的距离.

思路：先缩点，然后求最短路，floyd超时，所以用dij。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <cstdlib>#include <map>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll long longusing namespace std;const int maxn=505;const double INF=1000000000;struct node{    int v,w,next;}edge[maxn*maxn];int head[maxn],scc[maxn],G[maxn][maxn],dis[maxn];int low[maxn],dfn[maxn],stack[maxn],sum[maxn];bool ins[maxn],vis[maxn];int n,m,num,cnt,snum,top;void init(){    memset(head,-1,sizeof(head));    num=0;}void add(int u,int v,int w){    edge[num].v=v;    edge[num].w=w;    edge[num].next=head[u];    head[u]=num++;}void dfs(int u){    int x;    dfn[u]=low[u]=++cnt;    stack[top++]=u;    ins[u]=true;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(!dfn[v])        {            dfs(v);            low[u]=min(low[u],low[v]);        }        else if(ins[v]) low[u]=min(low[u],dfn[v]);    }    if(low[u]==dfn[u])    {        snum++;        do{            x=stack[--top];            ins[x]=false;            scc[x]=snum;        }while(x!=u);    }}void tarjan(){    memset(ins,false,sizeof(ins));    memset(dfn,0,sizeof(dfn));    snum=top=cnt=0;    for(int i=1;i<=n;i++)    if(!dfn[i]) dfs(i);}void make_graph(){    for(int i=1;i<=n;i++)    for(int j=1;j<=n;j++)    G[i][j]=INF;    for(int u=1;u<=n;u++)    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(scc[u]==scc[v]) G[u][v]=0;        else G[u][v]=edge[i].w;    }}void dij(int u,int v){    int mm,x;    memset(vis,false,sizeof(vis));    for(int i=1;i<=n;i++)    dis[i]=G[u][i];    dis[u]=0;    vis[u]=true;    for(int i=1;i<=n;i++)    {        mm=INF;        for(int j=1;j<=n;j++)        if(!vis[j]&&dis[j]<mm) mm=dis[x=j];        if(mm==INF) break;        vis[x]=true;        for(int j=1;j<=n;j++)        if(!vis[j]&&dis[j]>dis[x]+G[x][j])        dis[j]=dis[x]+G[x][j];    }    if(dis[v]==INF) printf("Nao e possivel entregar a carta\n");    else printf("%d\n",dis[v]);}int main(){    int k,a,b,c;   while(scanf("%d%d",&n,&m),n)   {       init();       while(m--)       {        scanf("%d%d%d",&a,&b,&c);        add(a,b,c);       }       tarjan();       make_graph();       scanf("%d",&k);       while(k--)       {           scanf("%d%d",&a,&b);           dij(a,b);       }       printf("\n");   }   return 0;}