HDU 2145 zz's Mysterious Present

zz's Mysterious Present

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 862    Accepted Submission(s): 174

Problem Description

There are m people in n cities, and they all want to attend the party which hold by zz. They set out at the same time, and they all will choose the best way they think, but due to someone take a ride, someone drive, and someone take a taxi, they have different speed. Can you find out who will get zz's mysterious present? The first one get the party will get the present . If there are several people get at the same time, the one who stay in the city which is farther from the city where is zz at begin will get the present. If there are several people get at the same time and the distance from the city he is at begin to the city where zz is, the one who has the larger number will get the present.

 

Input

The first line: three integers n, m and k. m is the total number of the people, and n is the total number of cities, and k is the number of the way.(0<n<=300, 0<m<=n, 0<k<5000)
The second line to the (k+1)th line: three integers a, b and c. There is a way from a to b, and the length of the way is c.(0<a,b<=n, 0<c<=100)
The (k+2)th line: one integer p(0<p<=n), p is the city where zz is.
The (k+3)th line: m integers. the ith people is at the place p[i] at begin.(0<p[i]<=n)
The (k+4)th line: m integers. the speed of the ith people is speed[i];(0<speed[i]<=100)
All the ways are directed.

 

Output

For each case, output the one who get the present in one line. If no one can get the present, output "No one".

 

Sample Input

3 1 31 2 21 3 32 3 1321

 

Sample Output

1

题目大意：m个人住在n个城市，他们都会选择最快的路去参加zz的聚会，已知m个人的起始点，速度和zz的地点，求出 最早到的人的序号，如果有同时到的，则取距离远的，如果距离相同则取序号大的.所有路是单向的 

思路：将终点当作起点然后一次Dijkstra，就可以求出所有点到终点的距离，因为对所有边做了反向处理，所以输入时也要反向输入.需要注意的是所有路是单项的.

#include<stdio.h>#define MAX 0xffffffint select[1000], dis[1000], map[1000][1000];void ShortPath(int s, int n) //dij算法模版{int i, j, k, min;for( i = 1; i <= n; i++)dis[i] = map[s][i];dis[s] = 0;select[s] = 1;for( i = 2; i <= n; i++){k = -1, min = MAX;for( j = 1; j <= n; j++)if(select[j] != 1 && dis[j] < min){k = j;min = dis[j];}if(k == -1)break;select[k] = 1;for( j = 1; j <= n; j++ )if( select[j] != 1 && dis[k] + map[k][j] < dis[j])dis[j] = dis[k] + map[k][j];}}int main(){int n, m, k, p, place[1000];int x, y, s, flag = 0, min;float speed[1000], result[1000];while( scanf("%d%d%d", &n, &m, &k) != EOF){int i, j;for( i = 1; i <= n; i++){dis[i] = MAX;select[i] = 0;for( j = 1; j <= n; j++){if(i == j)map[i][j] = 0;elsemap[i][j] = MAX;}}while(k--){scanf("%d%d%d", &x, &y, &s);  //反向并单向输入if(map[y][x] > s)map[y][x] = s;}scanf("%d", &p);ShortPath( p, n);for( i = 1; i <= m; i++)scanf("%d", &place[i]);for( i = 1; i <= m; i++)scanf("%f", &speed[i]);min = 1;for( i = 1; i <= m; i++)   //序号筛选{result[i] = dis[place[i]]*1.0 / speed[i];if(result[min] > result[i])min = i;else if(result[min] == result[i]){if(dis[place[min]] <= dis[place[i]])min = i;}}if( dis[place[min]] == MAX)printf("No one\n");elseprintf("%d\n", min);}return 0;}