[题解]SP703 SERVICE - Mobile Service_Done

设计状态F[i][j][k]代表完成任务i后，有一个员工在地点P[i]，其他两人分别在j和k两地。所需要的最小代价。

转移的方式：

分别考虑派遣i，j，k三人前往下一个需求地点，并更新状态。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define re register
#define GC getchar()
#define Clean(X,K) memset(X,K,sizeof(X))
#include <iostream>
#define Max(A,B) ((A)>(B)?(A):(B))
#define Min(A,B) ((A)<(B)?(A):(B))
using namespace std ;
int Qread () {
    int X =  ;
    char C = GC ;
    while (C > '' || C < '') C = GC ;
    while (C >='' && C <='') {
        X = X *  + C - '' ;
        C = GC ;
    }
    return X ;
}
const int Maxl =  , Maxn =  ,INF =  << ;
int C[Maxl][Maxn] , N , L , Times , Q[Maxn] = {}, F[Maxn][Maxl][Maxl];
int main () {
//    freopen ("SP703.in" , "r" , stdin) ;
    Times = Qread () ;
    while (Times -- ) {
        Clean (F , ) ;
        L = Qread () , N = Qread () ;
        for (re int i =  ; i <= L; ++ i) for (re int j =  ; j <= L; ++ j) C[i][j] = Qread () ;
        for (re int i =  ; i <= N; ++ i) Q[i] = Qread () ;
        Clean (F , 0x3f) ;
        F[][][] =  ;
        for (re int i =  ; i < N; ++ i) {
            for (re int j =  ; j <= L ; ++ j) {
                for (re int k =  ; k <= L; ++ k) {
                    if (Q[i] == j || Q[i] == k || j == k) continue ;
                    F[i + ][j][k] = Min (F[i + ][j][k] ,F[i][j][k] +  C[Q[i]][Q[i + ]]) ;
                    F[i + ][Q[i]][k] = Min (F[i + ][Q[i]][k] , F[i][j][k] + C[j][Q[i + ]]) ;
                    F[i + ][j][Q[i]] = Min (F[i + ][j][Q[i]] , F[i][j][k] + C[k][Q[i + ]]) ;
                }
            }
        }
        int Ans = INF ;
        for (re int i =  ; i <= L ; ++ i) {
            for (re int j =  ; j <= L; ++ j) {
                Ans = Min (Ans , F[N][i][j]) ;
            }
        }
        printf ("%d\n" , Ans) ;
    }
    fclose (stdin) , fclose (stdout) ;
    return  ;
}