poj 3114

Countries in War

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1725 Accepted: 550

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：一些间谍之间进行通信，同一个城市之间通信不需要花费时间。给出起点终点，问最小的时间花费。（给出一个有向带权图，在一个强连通分量的点之间距离设为了零，给出起点终点，求着两点之间的最短距离）。

思路：强连通分量+缩点+dij求最短路径。缩点之后构图，然后求这些点之间的距离，给定的起始点与终点，要判断在哪一个强连通分量中，然后求出即可。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define max_n 505#define max_e 250002#define inf 99999999int map[505][505],map1[505][505];int stack[max_n],top;//栈int isInStack[max_n];//是否在栈内int low[max_n],dfn[max_n],time;//点的low，dfn值；time从1开始int node_id;//int head[max_n],s_edge;//邻接表头  s_edge从1开始int gro_id[505];int vis[505],dis[505];struct Node{    int to;    int next;} edge[max_e];void init(){    top=0;    node_id=0;    memset(isInStack,0,sizeof(isInStack));    memset(low,0,sizeof(low));    memset(dfn,0,sizeof(dfn));    time=0;    memset(head,0,sizeof(head));    s_edge=0;    memset(edge,0,sizeof(edge));}void addedge(int u,int v){    s_edge++;    edge[s_edge].to=v;    edge[s_edge].next=head[u];    head[u]=s_edge;}int min(int a,int b){    if(a<b)return a;    else return b;}void taijan(int u){    //low值为u或u的子树能够追溯到得最早的栈中节点的次序号    stack[top++]=u;    isInStack[u]=1;    dfn[u]=++time; //记录点u出现的记录，并放在栈中    low[u]=time;    int e,v;    for(e=head[u]; e; e=edge[e].next) //如果是叶子节点，head[u]=0,edge[e].next=0;    {        v=edge[e].to;        if(!dfn[v])        {            taijan(v);            low[u]=min(low[u],low[v]);        }        else if(isInStack[v])            low[u]=min(low[u],dfn[v]);    }    int j;    if(dfn[u]==low[u])    {        node_id++;        while(j=stack[--top])        {            isInStack[j]=0;            gro_id[j]=node_id;            if(j==u)break;        }    }}void dij(int st,int end){    int k=st;    for(int i=1; i<=node_id; i++)    {        vis[i]=0;        dis[i]=map1[st][i];    }    vis[st]=1;    for(int i=1; i<node_id; i++)    {        int mmin=inf;        for(int j=1; j<=node_id; j++)        {            if(!vis[j]&&mmin>dis[j])            {                mmin=dis[j];                k=j;            }        }        vis[k]=1;        for(int j=1; j<=node_id; j++)            if(!vis[j]&&dis[j]>dis[k]+map1[k][j])                dis[j]=dis[k]+map1[k][j];    }    if(dis[end]==inf)        printf("Nao e possivel entregar a carta\n");    else        printf("%d\n",dis[end]);}int main(){    int N,M,A,B,C,num;    while(scanf("%d%d",&N,&M))    {        if(N==0&&M==0)break;        init();        for(int i=1; i<=N; i++)            for(int j=1; j<=N; j++)            {                if(i==j)                {                    map[i][j]=0;                    map1[i][j]=0;                }                else                {                    map[i][j]=inf;                    map1[i][j]=inf;                }            }        for(int i=0; i<M; i++)        {            scanf("%d%d%d",&A,&B,&C);            addedge(A,B);            if(map[A][B]>C)            map[A][B]=C;        }        for(int i=1;i<=N;i++)        if(!dfn[i])        taijan(i);        for(int i=1; i<=N; i++)        {            for(int j=1; j<=N; j++)                if(i!=j&&map[i][j]!=inf&&gro_id[i]!=gro_id[j])                {                    map1[gro_id[i]][gro_id[j]]=min(map[i][j],map1[gro_id[i]][gro_id[j]]);                }        }        scanf("%d",&num);        while(num--)        {            int st,end;            scanf("%d%d",&st,&end);            if(gro_id[st]==gro_id[end])                printf("0\n");            else                dij(gro_id[st],gro_id[end]);        }        printf("\n");    }    return 0;}/*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 2*/