博弈DP

每次取最左端或最右端的石子,谁取得多谁胜利

#include<iostream>
using namespace std;
const int N = 100;
int piles[N],dp[N][N][2];//0.表先手,1.表后手
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;++i)cin>>piles[i];
    //base case 
    for(int i=0;i<n;++i){
        dp[i][i][0]=piles[i];
        dp[i][i][1]=0;
    }
    //dp[i][j][0] i-j堆的先手情况 
    //dp[i][j][0]=piles[i]+dp[i+1][j][1]
    //dp[i][j][0]=piles[j]+dp[i][j-1][1]
    //dp[i][j][1]=dp[i+1][j][0]
    //dp[i][j][1]=dp[i][j-1][0];
    for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
            cout<<dp[i][j][0]<<" "<<dp[i][j][1]<<"------";
        }
        cout<<endl;
    }
    cout<<endl;
    for(int l=1;l<n;++l){
        for(int i=0;i<n;++i){
            int j=l+i;
            int left=piles[i]+dp[i+1][j][1];
            int right=piles[j]+dp[i][j-1][1];
            if(left>right){
                dp[i][j][0]=left;
                dp[i][j][1]=dp[i+1][j][0];
            }
            else{
                dp[i][j][0]=right;
                dp[i][j][1]=dp[i][j-1][0];
            }
        }
        for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
            cout<<dp[i][j][0]<<" "<<dp[i][j][1]<<"------";
        }
        cout<<endl;
    }
    cout<<endl;
    }
    cout<<dp[0][n-1][0]-dp[0][n-1][1]<<endl;
    return 0;
}
原文地址:https://www.cnblogs.com/clear-love/p/11335330.html