【PAT】1003. Emergency (25)

最新推荐文章于 2020-04-22 16:28:20 发布

原创最新推荐文章于 2020-04-22 16:28:20 发布 · 447 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm #DFS #最短路径

PAT 专栏收录该内容

107 篇文章

订阅专栏

本文介绍了一种寻找两点间最短路径及最大救援队伍数量的方法。通过DFS和Dijkstra+DFS两种方式，实现对特殊地图上救援路线的选择优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

分析：求两点之间最短路径的条数，并且找出经过最短路径所能携带的最大救援队伍数目。

方法一：采用DFS。

（参考了这篇博客：  http://blog.youkuaiyun.com/iaccepted/article/details/21451949#reply）

#include<iostream>
#include<string.h>
#include <climits>
using namespace std;
#define MAX 501
int weight[MAX]; // rescure teams @ each city
int map[MAX][MAX];
bool visited[MAX];
int N, M, cnt, tmpDist, maxWeight;
void init(int cityNum){
	int i, j;
	for(i=0; i<cityNum; i++){
		weight[i] = 0;
		for(j=0; j<MAX; j++){
			map[i][j] = map[j][i] = INT_MAX;
		}
	}
	memset(visited, false, sizeof(visited));
}

void dfs(int from, int to, int dist, int wei){
	if(from == to){
		if(dist < tmpDist){
			cnt = 1;
			tmpDist = dist;
			maxWeight = wei;			
		}else if(dist == tmpDist){
			++cnt;
			if(maxWeight < wei){
				maxWeight = wei;
			}
		}
		return;
	}
	if(dist>tmpDist) return; 

	for(int i=0; i<N; i++){
		if(!visited[i] && map[from][i]!=INT_MAX){
			visited[i] = true;
			dfs(i, to, dist+map[from][i], wei+weight[i]);
			visited[i] = false;
		}
	}
}


int main(){
	int from, to, i;
	int c1, c2, L;
	cin>>N>>M>>from>>to;
	init(N);
	cnt = 0;
	tmpDist = INT_MAX;
	maxWeight = 0;
	for(i=0; i<N; i++){
		cin>>weight[i];
	}
	for(i=0; i<M; i++){
		cin>>c1>>c2>>L;
		map[c1][c2] = map[c2][c1] = L;
	}
	dfs(from, to, 0, weight[from]);
	cout<<cnt<<" "<<maxWeight<<endl;
	return 0;
}

以上要注意的是如下这几行代码。

	if(!visited[i] && map[from][i]!=INT_MAX){
			visited[i] = true;
			dfs(i, to, dist+map[from][i], wei+weight[i]);
			visited[i] = false;
		}

之所以先让visited[i] = true 是为了防止逆向查找，即从A到B后，当从B开始查找时，不会再找到A。但是查完A所有的孩子节点后要将标记撤销，从而可以进行其他的查找。

方法二： Dijkstra + DFS

（参考： http://blog.youkuaiyun.com/abcjennifer/article/details/19410027）