PAT 解題報告 1003. Emergency (25)

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

題目描述：

抽象出來就是給你一張圖，給你起點終點，着到起點到終點的所有最短路徑，每個點上有一個數，代表這個點上的搜救隊的數量，要求輸出這些最短路徑中你能召集到的最大的搜救隊的數量。

算法分析：

思路1、dijsktra

簡單題，最短路徑算法(e.g. Dijastra)，但是要稍微變化一下，要記錄所有的的最短路徑，所以從終點回溯上去可能有多於一個的parent。那么就在每個node上面掛一個 list好了，記錄所有的parents. 得到后可以從終點DFS到起點，記錄下每條最短路徑上的搜救隊的數量和。保存最大的那個輸出即可。

思路2、DFS

注意點：

這個題目其實就是最短路徑的小小變形，注意最原始的最短路徑算法通過記錄每個node最短路徑上的parent來保存一條這樣的最短路徑，這里話就 是相當於，要多點判斷，當新的路徑和當前路徑相同長度時，也要記錄下來，但是后面如果找到更短的，那么之前的路徑就要清空，記錄路徑的代碼稍微注意一點就 沒問題了。

Dijkstra

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

#define INF 0x3f3f3f3f
#define MX 501

int mp[MX][MX];
int v[MX];
int dist[MX];
int amount[MX];
int teams[MX];
int pathcount[MX];
int N,M,start,en;

void dijkstra(int s){
    amount[s] = teams[s];
    dist[s] = 0;
    pathcount[s] = 1;

    while (1){
        int u, dmin=INF;
        for (int i=0; i<N; i++){
            if (v[i]==0 && dist[i]<dmin){
                dmin = dist[i];
                u = i;
            }
        }
        if (dmin==INF || u==en) break;
        v[u] = 1;
        for (int i=0; i<N; i++){
            if(v[i]==0){
                if (dist[i] > dist[u] + mp[u][i]){
                    dist[i] = dist[u] + mp[u][i];
                    amount[i] = amount[u] + teams[i];
                    pathcount[i] = pathcount[u];
                }else if (dist[i] == dist[u] + mp[u][i]){
                    pathcount[i] += pathcount[u];
                    if (amount[i] < amount[u] + teams[i])
                        amount[i] = amount[u] + teams[i];
                }
            }
        }
    }
}

int main()
{
    scanf("%d%d%d%d", &N,&M,&start,&en);
    for (int i=0; i<N; i++)
    {
        scanf("%d", &teams[i]);
    }
    for (int i=0; i<N; i++)
    {
        dist[i] = INF;
        for (int j=0; j<N; j++)
            mp[i][j] = INF;
    }
    for (int i=0; i<M; i++)
    {
        int c1, c2, L;
        scanf("%d%d%d", &c1,&c2,&L);
        mp[c1][c2] = mp[c2][c1] = L;
    }

    dijkstra(start);
    printf("%d %d", pathcount[en], amount[en]);

    return 0;
}

 

DFS

#include<cstdio>
#include<cstring>

#define INF 0x7FFFFF
int u[502]={0};
int teams[502]={0};
int dist[502];
int mp[502][502];
int n,m,st,en;
int shortNum=0,maxteam=0,mindist=INF;

void dfs(int s,int dis,int team){//到達S結點時的距離，teams
    if(s==en){
        if(dis<mindist){
            mindist=dis;
            shortNum=1;
            maxteam=team;
        }else if(dis==mindist){
            shortNum++;
            if(team>maxteam) maxteam=team;
        }
        return;
    }
    u[s]=1;
    for(int i=0;i<n;i++){
        if(u[i]==0 && mp[s][i]>0){
            dfs(i,dis+mp[s][i],team+teams[i]);
        }
    }
    u[s]=0;
}

int main(){
    freopen("in.txt","r",stdin);
    int i;
    scanf("%d%d%d%d",&n,&m,&st,&en);
    for(i=0;i<n;i++) scanf("%d",&teams[i]);
    memset(mp,-1,sizeof(mp));
    for(i=0;i<m;i++){
        int t1,t2,dis;
        scanf("%d%d%d",&t1,&t2,&dis);
        mp[t1][t2]=mp[t2][t1]=dis;
    }

    //u[st]=1;
    dfs(st,0,teams[st]);

    printf("%d %d\n",shortNum,maxteam);
    return 0;
}