SPOJ 703 SERVICE - Mobile Service 题解

题面

好题啊！~

设f[i][j][k][l]表示已经处理完前i个请求后，a在j，b在k，c在l的最小值是多少；

那么f[i][p[i]][k][l]=min(f[i][p[i]][k][l],f[i-1][j][k][l]+c(j,p[i]));

f[i][j][p[i]][l]=min(f[i][j][p[i]][l],f[i-1][j][k][l]+c(k,p[i]));

f[i][j][k][p[i]]=min(f[i][j][k][p[i]],f[i-1][j][k][l]+c(l,p[i]));

但以上的复杂度卡死你都过不去，所以优化状态；

因为对于i，肯定有一个人处于p[i];所以把p[i]舍弃

f[i+1][j][k]=min(f[i+1][j][k],f[i][j][k]+a[p[i]][p[i+1]]);
f[i+1][p[i]][k]=min(f[i+1][p[i]][k],f[i][j][k]+a[j][p[i+1]]);
f[i+1][j][p[i]]=min(f[i+1][j][p[i]],f[i][j][k]+a[k][p[i+1]]);

注意p[0]=3;

f[0][1][2]=0;

#include <bits/stdc++.h>
using namespace std;
int a[][];
int f[][][];
int p[];
int main()
{
    int t;
    cin>>t;
    while(t--){
        int l,n;
        cin>>l>>n;
        for(register int i=;i<=l;i++){
            for(register int j=;j<=l;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=;i<=n;i++){
            scanf("%d",&p[i]);
        }
        memset(f,0x3f,sizeof(f));
        f[][][]=;
        p[]=;
        for(register int i=;i<n;i++){
            for(register int j=;j<=l;j++){
                for(register int k=;k<=l;k++){
                    if(j==p[i]) continue;
                    if(k==p[i]) continue;
                    if(j==k) continue;
                    if(j!=p[i+]&&k!=p[i+]) f[i+][j][k]=min(f[i+][j][k],f[i][j][k]+a[p[i]][p[i+]]);
                    if(p[i]!=p[i+]&&k!=p[i+]) f[i+][p[i]][k]=min(f[i+][p[i]][k],f[i][j][k]+a[j][p[i+]]);
                    if(j!=p[i+]&&p[i]!=p[i+]) f[i+][j][p[i]]=min(f[i+][j][p[i]],f[i][j][k]+a[k][p[i+]]);
                }
            }
        }
        int minn=INT_MAX;
        for(int i=;i<=l;i++){
            for(int j=;j<=l;j++){
                minn=min(minn,f[n][i][j]);
            }
        }
        cout<<minn<<endl;
    }
}