Seaside(hdu3665,floyd)
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http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29015#problem/C
http://acm.hdu.edu.cn/showproblem.php?pid=3665
Seaside
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 957 Accepted Submission(s): 681
Problem Description
XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ¡¯0¡¯. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered ¡¯0¡¯, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.
Input
There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers SMi and LMi, which means that the distance between the i-th town and the SMi town is LMi.
Output
Each case takes one line, print the shortest length that XiaoY reach seaside.
Sample Input
5
1 0
1 1
2 0
2 3
3 1
1 1
4 100
0 1
0 1
Sample Output
2
Source
2010 Asia Regional Harbin
Recommend
lcy
解析:
思路:
有向图,求单源最短路,起点是1终点不定,求1到达海边的最短路
232 KB 0 ms C++ 1051 B
*/
#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#include <iostream>using namespace std;const int maxn=20;const int inf=100000001;int d[maxn][maxn];int n;struct node{int cnt;int flag;}p[maxn];void floyd(){for(int k=0;k<n;k++) for(int i=0;i<n;i++) for(int j=0;j<n;j++) { if(d[i][k]<inf&&d[k][j]<inf) { if(d[i][j]>d[i][k]+d[k][j]) {d[i][j]=d[i][k]+d[k][j]; } } }}int main(){ int i,j,ok; while(scanf("%d",&n)!=EOF) { int u,w; for(i=0;i<n;i++)for(j=0;j<n;j++){if(i==j)d[i][j]=0;elsed[i][j]=inf;} for(i=0;i<n;i++) { scanf("%d%d",&p[i].cnt,&p[i].flag); for(j=0;j<p[i].cnt;j++) { scanf("%d%d",&u,&w); if(w<d[i][u]) d[i][u]=w; } } floyd(); int ans=inf; for(i=0;i<n;i++) { if(ans>d[0][i]&&p[i].flag==1) ans=d[0][i]; } printf("%d\n",ans); } // system("pause"); return 0;}
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