Problem Description

The ocean is a treasure house of resources and the development of human society comes to depend more and more on it. In order to develop and utilize marine resources, it is necessary to build mining stations on the sea. However, due to seabed mineral resources, the radio signal in the sea is often so weak that not all the mining stations can carry out direct communication. However communication is indispensable, every two mining stations must be able to communicate with each other (either directly or through other one or more mining stations). To meet the need of transporting the exploited resources up to the land to get put into use, there build n ports correspondently along the coast and every port can communicate with one or more mining stations directly.

Due to the fact that some mining stations can not communicate with each other directly, for the safety of the navigation for ships, ships are only allowed to sail between mining stations which can communicate with each other directly. 

The mining is arduous and people do this job need proper rest (that is, to allow the ship to return to the port). But what a coincidence! This time, n vessels for mining take their turns to take a rest at the same time. They are scattered in different stations and now they have to go back to the port, in addition, a port can only accommodate one vessel. Now all the vessels will start to return, how to choose their navigation routes to make the total sum of their sailing routes minimal. 

Notice that once the ship entered the port, it will not come out!

Input

There are several test cases. Every test case begins with four integers in one line, n (1 = <n <= 100), m (n <= m <= 200), k and p. n indicates n vessels and n ports, m indicates m mining stations, k indicates k edges, each edge corresponding to the link between a mining station and another one, p indicates p edges, each edge indicating the link between a port and a mining station. The following line is n integers, each one indicating one station that one vessel belongs to. Then there follows k lines, each line including 3 integers a, b and c, indicating the fact that there exists direct communication between mining stations a and b and the distance between them is c. Finally, there follows another p lines, each line including 3 integers d, e and f, indicating the fact that there exists direct communication between port d and mining station e and the distance between them is f. In addition, mining stations are represented by numbers from 1 to m, and ports 1 to n. Input is terminated by end of file.

Output

Each test case outputs the minimal total sum of their sailing routes.

Sample Input

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

Sample Output

13

————————————————————————————————————————————————————

題意：有一 n 個港口 m 個油田的無向圖，並給出了圖中所有的邊與權值，現在給出 n 個船所在的油田編號，要讓這 n 條船每一條頭回到一個港口中，且每個港口只能容納一條船，求這 n 條船行走的距離總和的最小值

思路：每條船到任意港口均有一個最短距離，首先要求出最短距離，然後建立一個二分圖，左點集為 n 個港口，右點集為 n 條船，若第 j 條船到第 i 個港口的距離為 x，則連一條左 i 右 j 的邊，由於要求的是最小值，因此需要對所有邊取負再使用 KM 演算法，最後得出的最優匹配取負，即為答案 

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 1001
#define LL long long
using namespace std;
int n;
int G[N][N];
int Lx[N],Ly[N];
bool visX[N],visY[N];
int linkX[N],linkY[N];
bool dfs(int x){
    visX[x]=true;
    for(int y=1;y<=n;y++){
        if(!visY[y]){
            int temp=Lx[x]+Ly[y]-G[x][y];
            if(temp==0){
                visY[y]=true;
                if(linkY[y]==-1 || dfs(linkY[y])){
                    linkX[x]=y;
                    linkY[y]=x;
                    return true;
                }
            }
        }
    }
    return false;
}
void update(){
    int minn=INF;
    for(int i=1;i<=n;i++)
        if(visX[i])
            for(int j=1;j<=n;j++)
                if(!visY[j])
                    minn=min(minn,Lx[i]+Ly[j]-G[i][j]);

    for(int i=1;i<=n;i++)
        if(visX[i])
            Lx[i]-=minn;

    for(int i=1;i<=n;i++)
        if(visY[i])
            Ly[i]+=minn;
}
int KM(){
    memset(linkX,-1,sizeof(linkX));
    memset(linkY,-1,sizeof(linkY));
    memset(Lx,0,sizeof(Lx));
    memset(Ly,0,sizeof(Ly));

    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            Lx[i]=max(Lx[i],G[i][j]);

    for(int i=1;i<=n;i++){
        while(true){
            memset(visX,false,sizeof(visX));
            memset(visY,false,sizeof(visY));

            if(dfs(i))
                break;
            else
                update();
        }
    }

    int ans=0;
    for(int i=1;i<=n;i++)
        if(linkY[i]!=-1)
            ans+=G[linkY[i]][i];

    return ans;
}
int id[N];
int dis[N][N];
int main(){
    int m,k,p;
    while(scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF&&(n+m+k+p)){
        for(int i=1;i<=n;i++)
            scanf("%d",&id[i]);

        memset(dis,INF,sizeof(dis));
        for(int i=1;i<=n+m;i++)
            for(int j=1;j<=n+m;j++)
                if(i==j)
                    dis[i][j]=0;

        while(k--){
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            dis[x+n][y+n]=w;
            dis[y+n][x+n]=w;
        }

        while(p--){
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            dis[x][y+n]=w;
        }

        for(int k=1;k<=n+m;k++)
            for(int i=1;i<=n+m;i++)
                for(int j=1;j<=n+m;j++)
                    if(dis[i][k]<INF && dis[k][j]<INF)
                        dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);

        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                G[i][j]=-dis[i][id[j]+n];

        printf("%d\n",-KM());
    }
    return 0;
}