1003

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1003. Emergency (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

Sample Output

2 4

题意：

你是救援队队长，带领你的队伍去救人，每经过一个城市就可以带着这座城市的救援队一起走，给你目的地，城市和城市之间路的距离，和每个城市有的队伍数量，求最短路径和能得到的最多队伍数

思路：

神搜，遍历每一个从该点出发到终点的可能路径，比较得出最小的

递归的结束条件：到了终点

到了终点发生的改变：如果走过的路径是最小路径，team数变成新的team数；如果走过的路径和最小的相等，team数等于更多的

ps：用max保存最多的team数，而当前的team数，是传参而得的

#include<stdio.h>
#define MAXVER 501
#define MAXEDG 12501
typedef struct{
	int vex[MAXVER];//teams in every city
	int arc[MAXVER][MAXVER];//length of road
	int N, M;
}Mgraph;

Mgraph g;
int visited[MAXVER], c1, c2, count, max, min = 66235;

void read(){
	int i, j, r, c;
	scanf("%d%d%d%d", &(g.N), &(g.M), &c1, &c2);
	for(i = 0; i < g.N; i++)//the number of teams in i city
		scanf("%d", &g.vex[i]);
	for(i = 0; i < MAXVER; i++)//init, make every roads be 0
		for(j = 0; j < MAXVER; j++){
			g.arc[i][j] = 0;
		}
		for(i = 0; i < g.M; i++){//length of the road
			scanf("%d%d%d", &r, &c, &j);
			g.arc[r][c] = g.arc[c][r] = j;    
		}
}

void dfs(int now, int right, int team)
{
	int i;
	if(right > min)
		return;
	if(now == c2)
	{
		if(right == min){
			count ++;
			if(team >= max)
				max = team;
		}
		if(right < min){
			min = right;
			count = 1;
			max = team;
		}
		return ;
	}
	for(i = 0; i < g.N; i++){
		if(!visited[i] && g.arc[now][i] > 0)	{
			visited[i] = 1;
			dfs(i, right + g.arc[now][i], team + g.vex[i]);
			visited[i] = 0;
		}
	}
}

int main(){

	//freopen("in.txt", "r", stdin);
	read();
//	if(c1 == c2)
//		printf("1 g.vex[c1]");
//	else{
		visited[c1] = 1;
		dfs(c1, 0, g.vex[c1]);
		printf("%d %d", count, max); 
//	}
	return 0;
}