PAT 1003

Posted 2020-10-28 chlewn

1003. Emergency (25)

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

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

---

图典型的最短路径问题。从城市C1出发，目的地C2，第一要求是路径最短，其次是路径上各点上的man之和最大，
输出的是最短路径的条数和man之和最大数

思路：

先构建邻接矩阵edges，这里需要给edge初始一个无穷大数值，忘记了c怎么给一个数组每个元素设初值的函数调用，所以将edge定义为一个结构体，有两个成员变量，一个表示是否可到达，一个表示距离。

另外设置一个marked数组记录每个城市点是否被访问过

struct edge{
    bool can_reach;
    int length;
    edge(){ can_reach = false; }
};
edge edges[505][505];
bool marked[505] = {false};

具体代码：

#include <iostream>
#include <algorithm>
#include <climits>
using namespace std;
int city_hands[505] = {0};
struct edge{
    bool can_reach;
    int length;
    edge(){ can_reach = false; }
};
edge edges[505][505];
bool marked[505] = {false};
int min_path = INT_MAX;
int max_hands = INT_MIN;
int num = 0;
int N_cities,M_roads,C1,C2;
void dfs(int cur_city,int len,int hands)
{
    if(min_path<len)    return;
    if(cur_city==C2){
        if(min_path>len){
            num = 1;
            min_path = len;
            max_hands = hands;
        }
        else if(min_path==len){
            num ++;
            if(max_hands<hands) max_hands = hands;
        }
        return;
    }
    for(int i=0;i<N_cities;i++){
        if(marked[i]==false&&edges[cur_city][i].can_reach==true){
            marked[i] = true;
            dfs(i,len+edges[cur_city][i].length,hands+city_hands[i]);
            marked[i] = false;
        }
    }
    return;
}
int main()
{
    cin >> N_cities >> M_roads >> C1 >> C2;
    for(int i=0;i<N_cities;i++) cin>>city_hands[i];
    for(int i=0;i<M_roads;i++){
        int a,b,l;
        cin>>a>>b>>l;
        edges[a][b].can_reach = true;
        edges[a][b].length = l;
        edges[b][a].can_reach = true;
        edges[b][a].length = l;
    }
    marked[C1] = true;
    dfs(C1,0,city_hands[C1]);
    cout<<num<<" "<<max_hands<<endl;
    return 0;
}