poj 3615 Cow Hurdles
2013-05-22 23:43
465 查看
题目大意:很多单向路上有一个Hurdles,给你Hurdles的高度,求出走到目标地点所需要的最小的跳高的能力。
#include<cstdio> #include<iostream> int map[400][400]; int dis[400][400]; const int INF=1<<28; int n,m,t; int main() { while(scanf("%d%d%d",&n,&m,&t)!=EOF) { int i,j; for(i=1;i<=n;i++) for(j=1;j<=n;j++) map[i][j]=INF; //初始化 for(i=1;i<=m;i++) { int a,b,c; scanf("%d%d%d",&a,&b,&c); if(c<map[a][b])map[a][b]=c; //建图 } for(i=1;i<=n;i++) for(j=1;j<=n;j++) { dis[i][j]=map[i][j]; dis[i][i]=0; } int k; for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) { int max=dis[i][k]<dis[k][j]?dis[k][j]:dis[i][k]; dis[i][j]=dis[i][j] > max ? max:dis[i][j]; } for(i=1;i<=t;i++) { int s,e; scanf("%d%d",&s,&e); if(dis[s][e]!=INF)printf("%d\n",dis[s][e]); else printf("-1\n"); } } return 0; }
相关文章推荐
- POJ 3615 Cow Hurdles(Floyd)
- 【POJ 3615】Cow Hurdles
- Poj 3615 Cow Hurdles【Floyd】
- POJ--3615|Cow Hurdles
- poj 3615 Cow Hurdles
- POJ 3615 Cow Hurdles ,floyd 修改 , UVa 10048 Audiophobia
- poj 3615 Cow Hurdles
- POJ3615 Cow Hurdles【Floyd】
- POJ 3615 Cow Hurdles(Folyd变形)
- poj 3615 :Cow Hurdles (floyd)----很好
- poj 3615 Cow Hurdles(floyd)
- POJ 3615 Cow Hurdles
- POJ 3615 Cow Hurdles (Floyd算法)
- poj 3615 Cow Hurdles
- POJ 3615 Cow Hurdles(Floyd算法)
- POJ 3615 Cow Hurdles
- poj 3615 Cow Hurdles
- POJ 3615 A - Cow Hurdles 弗洛伊德算法
- poj 3615 Cow Hurdles
- POJ - 3615 Cow Hurdles (Floyd变形)