2387 poj Til the Cows Come Home【dijkstra,经典&&基础】
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Til the Cows Come Home
Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 37068 Accepted: 12622Sample Input
5 51 2 202 3 303 4 204 5 201 5 100
Sample Output
90
题目大意:
给两个数 T , M 。T表示在接下来有T条边【即T行数据】,N表示有N个点,然后输入数据【注意1,重边时,取更短的】【注意2,数组开大点,笔者应为只开到100+,re了3次】,然后输出点 1 到点 N 的最小距离。
#include <cstdio>#include <cstring>#define INF 1000000using namespace std;int t, n;int dis[1001], vis[1001];int map[1001][1001];//数组开到1001就够了,之前笔者只开到 101 ,所以就re了3次 void Dijsktra() {for(int i = 1; i <= n; i++) {vis[i] = 0;dis[i] = INF;}dis[1] = 0;while(1) {int v = -1;for(int i = 1; i <= n; i++)if(!vis[i] && (v == -1 || dis[v] > dis[i]))v = i;if(v == -1)//注意这句,没了就成死循环了 break;vis[v] = 1;for(int i = 1; i <= n; i++)if(dis[i] > dis[v] + map[v][i])dis[i] = dis[v] + map[v][i];}printf("%d\n", dis[n]);}int main() {int a, b, c;while(~scanf("%d %d", &t, &n)) {//注意T为边数,N为点数 for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)map[i][j] = (i == j ? 0 : INF);for(int i = 0; i < t; i++) {scanf("%d %d %d", &a, &b, &c);if(map[a][b] > c)//重边取短的 map[a][b] = map[b][a] = c;}Dijsktra();}return 0;}
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