SPOJ 703 SERVICE - Mobile Service 题解

Posted 2022-09-28 kamimxr

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[1010][1010];
int f[1010][210][210];
int p[10010];
int main()

    int t;
    cin>>t;
    while(t--)
        int l,n;
        cin>>l>>n;
        for(register int i=1;i<=l;i++)
            for(register int j=1;j<=l;j++)
                scanf("%d",&a[i][j]);
            
        
        for(int i=1;i<=n;i++)
            scanf("%d",&p[i]);
        
        memset(f,0x3f,sizeof(f));
        f[0][1][2]=0;
        p[0]=3;
        for(register int i=0;i<n;i++)
            for(register int j=1;j<=l;j++)
                for(register int k=1;k<=l;k++)
                    if(j==p[i]) continue;
                    if(k==p[i]) continue;
                    if(j==k) continue;
                    if(j!=p[i+1]&&k!=p[i+1]) f[i+1][j][k]=min(f[i+1][j][k],f[i][j][k]+a[p[i]][p[i+1]]);
                    if(p[i]!=p[i+1]&&k!=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]]);
                    if(j!=p[i+1]&&p[i]!=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]]);
                
            
        
        int minn=INT_MAX;
        for(int i=1;i<=l;i++)
            for(int j=1;j<=l;j++)
                minn=min(minn,f[n][i][j]);
            
        
        cout<<minn<<endl;