Dijkstra-POJ-2387-Til the Cows Come Home
Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu
Description
Input
Output
Sample Input
Sample Output
Hint
Description
现在，已知起点和终点，请你计算出要从起点到终点，最短需要行走多少距离。
Input
每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000)，分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0～N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N)，分别代表起点和终点。
Output
Sample Input
3 3 0 1 1 0 2 3 1 2 1 0 2 3 1 0 1 1 1 2
Sample Output
2 -1
Hint
Description
Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
Input
* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
Output
Sample Input
5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100
Sample Output
90
Hint
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
但是要注意的是数据输入时的判断，可能会有相同两点多条路径，一定要保留最短的那条的数据。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Inf 0x3fffffff
using namespace std;
int mp[1005][1005];
int dis[1005];
void dijkstra(int n)
{int vis[1001]={0};int minn,i,j,k;vis[1]=1;for(i=1;i<n;++i){minn=Inf;k=1;for(j=1;j<=n;++j){if(!vis[j]&&minn>dis[j]){minn=dis[j];k=j;}}vis[k]=1;for(j=1;j<=n;++j){if(!vis[j]&&dis[j]>dis[k]+mp[k][j])dis[j]=dis[k]+mp[k][j];}}printf("%d\n",dis[n]);
}int main()
{int t,n,i,j,s,t,val;while(scanf("%d%d",&t,&n)!=EOF){for(i=1;i<=n;++i){mp[i][i]=0;for(j=1;j<i;++j)mp[i][j]=mp[j][i]=Inf;}for(i=1;i<=t;++i){scanf("%d%d%d",&s,&t,&val);if(val<mp[s][t])mp[s][t]=mp[t][s]=val;}for(i=1;i<=n;++i)dis[i]=mp[1][i];dijkstra(n);}return 0;
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