1003 Emergency (25)（25 point(s)）

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problem

1003 Emergency (25)（25 point(s)）
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

anwser

Dijkstra 解法
#include<bits/stdc++.h>

#define INF 0x3f3f3f3f
#define Max 511

int N, M, C1, C2;
int Rescue[Max], Map[Max][Max], Dis[Max], Pre[Max], W[Max], Diff[Max];
bool Vis[Max] = {false};

void Dijkstra(int s){
    memset(Dis, INF, sizeof(Dis));
    memset(W, 0, sizeof(W));
    memset(Diff, 0, sizeof(Diff));
    
    Dis[s] = 0;
    W[s] = Rescue[s];
    Diff[s] = 1;
    for(int i = 0; i < N; i++) Pre[i] = i;
    
    for(int i = 0; i < N; i++){
        int u = 0, minn = INF;
        for(int j = 0; j < N; j++){
            if(!Vis[j] && Dis[j] < minn){
                u = j;
                minn = Dis[j];
            }
        }
        
        if(u == C2 || minn == INF) return;
        Vis[u] = true;
        
        for(int v = 0; v < N; v++) {
            if(!Vis[v]) {
                if(Dis[u] + Map[u][v] < Dis[v]){
                    Dis[v] = Dis[u] + Map[u][v];
//                  Pre[v] = u;
//              }
                    W[v] = W[u] + Rescue[v];
                    Diff[v] = Diff[u];
                }else if (Dis[u] + Map[u][v] == Dis[v]){
                    Diff[v] += Diff[u];
                    if(W[u] + Rescue[v] > W[v]){
                        W[v] = W[u] + Rescue[v];
//                      Pre[v] = u;
                    }
                }
                
            }
        }
    }
}

int main(){
//  freopen("test.txt", "r", stdin);
    
    memset(Map, INF, sizeof(Map));
    
    std::cin>>N>>M>>C1>>C2;
    for(int i = 0; i < N; i++){
        std::cin>>Rescue[i];
    }
    
    for(int i = 0; i < M; i++){
        int c1, c2, L;
        std::cin>>c1>>c2>>L;
        Map[c1][c2] = Map[c2][c1] = L;
    }
    
    Dijkstra(C1);
    
    std::cout<<Diff[C2]<<" "<<W[C2];
    
    return 0;
}

/*
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
*/

DFS解法

未完待续

experience

• 注意审题，求的不是最短路，是最短路的不同路条数。
• 这个图不是单向图，是双向图。
• Dijkstra算法以及其变种需要熟悉。
  单词复习
• scattered 分散的

1003 Emergency (25)（25 point(s)）

标签：def   code   man   esc   least   jks   one   stdin   不同   

原文地址：https://www.cnblogs.com/yoyo-sincerely/p/9251529.html