POJ3114(缩点+最短路)
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Countries in War
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 to N) 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 city X 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 be K 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 city O 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. The I-th line should contain an integer M, the minimum time, in hours, to send a letter in the I-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 5 1 2 5 2 1 10 3 4 8 4 3 7 2 3 6 5 1 2 1 3 1 4 4 3 4 1 3 3 1 2 10 2 3 1 3 2 1 3 1 3 3 1 3 2 0 0 Sample Output 0 6 6 0 Nao e possivel entregar a carta 10 Nao e possivel entregar a carta 0 Source
South America 2006, Brazil Subregion
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题意：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：UV在一个强连通分量时肯定t=0，不在时缩点重新构图，求最短路即可。
PS： Floyd先写了一遍；T掉了，结果spfa跑了一遍，关键要重新构图，要开新的结构体，写起来稍微麻烦点，最后用了dijkstra，速度还不错，300+ms，debug好长时间......
/****************************
* author:crazy_石头
* date：2014/01/18
* algorithm:tarjan+dijkstra
* Pro：POJ 3114
***************************/
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
#define INF 1<<29
#define eps 1e-8
#define A system("pause")
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
const int maxn=2000;
const int maxm=20;
struct edge
{
int from,to,w,next;
}e[maxn*maxn];
int head[maxn],dfn[maxn],low[maxn],instack[maxn],cnt,index,n,m,stack[maxn],belong[maxn],Scc,map[maxn][maxn],top,d[maxn],vis[maxn];
inline void addedge(int u,int v,int w)
{
e[cnt].from=u;
e[cnt].to=v;
e[cnt].w=w;
e[cnt].next=head[u];
head[u]=cnt++;
}
inline void tarjan(int u)
{
low[u]=dfn[u]=++index;
stack[top++]=u;
instack[u]=true;
int v;
for(int i=head[u];~i;i=e[i].next)
{
v=e[i].to;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(instack[v])
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
++Scc;
do
{
v=stack[--top];
belong[v]=Scc;//标记连通分量；
instack[v]=false;
}while(u!=v);
}
}
inline void init()
{
ms(head,-1),cnt=0;
}
inline void solve()
{
index=top=Scc=0;
ms(low,0),ms(dfn,0),ms(instack,0);
rep(i,1,n) rep(j,1,n) i==j?map[i][j]=0:map[i][j]=INF;
rep(i,1,n)
if(!dfn[i]) tarjan(i);
int u,v;
rep(i,0,cnt-1)
{
u=e[i].from,v=e[i].to;
if(belong[u]!=belong[v]) map[belong[u]][belong[v]]=min(map[belong[u]][belong[v]],e[i].w);
}
rep(i,1,Scc) map[i][i]=0;//同一个连通分量距离为0；
}
inline int dijkstra(int s,int t)
{
ms(vis,0);
rep(i,1,n) d[i]=map[s][i];
d[s]=0;
rep(i,1,n)
{
int k;
int minm=INF;
rep(j,1,n)
{
if(!vis[j]&&d[j]<minm)
minm=d[j],k=j;
}
vis[k]=1;
rep(j,1,n)
d[j]=min(d[j],d[k]+map[k][j]);
}
return d[t];
}
int main()
{
int test,u,v,w;;
while(~scanf("%d%d",&n,&m),n)
{
init();
while(m--)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
}
solve();
scanf("%d",&test);
while(test--)
{
scanf("%d%d",&u,&v);
u=belong[u],v=belong[v];
int ans=dijkstra(u,v);
if(ans!=INF) printf("%d\n",ans);
else printf ("Nao e possivel entregar a carta\n");
}
printf("\n");
}
return 0;
}