PAT 1003 Emergency

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最新推荐文章于 2025-04-28 20:01:17 发布

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本文介绍了一种使用深度优先搜索（DFS）算法解决城市间紧急救援路线规划与资源调度的问题。通过遍历地图上的城市和道路，寻找从当前城市到目标城市的最短路径，并在最短路径上尽可能多地召集救援队伍。该方法适用于数据量较小的情况，通过剪枝策略提高效率。

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题目

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 Specification:

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, C1and 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 Specification:

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:

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output:

2 4

解答：

简单的说，就是只使用dfs来遍历整张图，从而获得最短路的条数和在最短路的条件下可以召集到的最多的人手。当然直接dfs是有风险的，但是这道题的数据量告诉我们可以瞎搞，所以dfs加简单的剪枝就足以解决问题。
易错点：当最小路径发生变化之后，对于原来获得数据都要进行重新的统计。

代码：

#include<iostream>
#include<cstdio>
using namespace std;

int val[505][505]={0},vis[505]={0};
int c1,c2,n,m,rec[505],road=0,mins=0x7fffffff,maxs=0;

void dfs(int x,int res,int res1){
    if(res>mins) return;
    if(x==c2&&res<=mins) {//这个地方其实写的太啰嗦，可以合并掉，另外注意不要有多余的输出，血的教训！！！
        if(res<mins) maxs = res1;
        else maxs=(maxs<res1?res1:maxs);
        if(res<mins)road=1;
        else road++;
        mins=res;
        return;
    }
    for(int i=0;i<n;++i){
        if(!vis[i]&&val[x][i]){
            vis[i] = 1;
            dfs(i,res+val[x][i],res1+rec[i]);
            vis[i] = 0;
        }
    }
}

int main(){
    scanf("%d%d%d%d",&n,&m,&c1,&c2);
    for(int i=0;i<n;++i) scanf("%d",&rec[i]);
    for(int i=0;i<m;++i) {
        int f,t,v;
        scanf("%d%d%d",&f,&t,&v);
        val[t][f] = v;
        val[f][t] = v;
    }
    vis[c1] = 1;
    dfs(c1,0,rec[c1]);
    printf("%d %d\n",road,maxs);
    return 0;
}