CSP认证 201803-4 棋局评估(极大极小值搜索)

题目连接：http://118.190.20.162/view.page?gpid=T70c++

题目大意：给一个3*3棋盘，问按照最优策略下，若是1能赢输出赢后剩余未下的格子数+1,2能赢输出赢后负的剩下未下的格子数-1，平局输出0spa

题目思路：3*3很小，直接暴力全部状况，先手下尽量想让值高，反手下尽量想让值低，因此只用在全部可能中尽量取利于本身的状况便可code

如下是代码：
 get

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
int mp[5][5];
int solve(int p){
    int num=0;
    rep(i,1,3){
        rep(j,1,3){
            if(!mp[i][j])num++;
        }
    }
    if(p==1)return num+1;
    else return -num-1;
}
int check(){
    int num1=0,num2=0,p=0;
    rep(i,1,3){
        num1=0,num2=0;
        rep(j,1,3){
            if(mp[i][j]==1)num1++;
            if(mp[i][j]==2)num2++;
        }
        if(num1==3){
            p=1;
            break;
        }
        if(num2==3){
            p=2;
            break;
        }
    }
    if(p)return solve(p);
    rep(j,1,3){
        num1=0,num2=0;
        rep(i,1,3){
            if(mp[i][j]==1)num1++;
            if(mp[i][j]==2)num2++;
        }
        if(num1==3){
            p=1;
            break;
        }
        if(num2==3){
            p=2;
            break;
        }
    }
    if(p)return solve(p);
    num1=0,num2=0;
    rep(i,1,3){
        if(mp[i][i]==1)num1++;
        if(mp[i][i]==2)num2++;
    }
    if(num1==3)return solve(1);
    if(num2==3)return solve(2);
    num1=0,num2=0;
    rep(i,1,3){
        if(mp[i][4-i]==1)num1++;
        if(mp[i][4-i]==2)num2++;
    }
    if(num1==3)return solve(1);
    if(num2==3)return solve(2);
    int flag=0;
    rep(i,1,3){
        rep(j,1,3){
            if(!mp[i][j]){
                flag=1;
                break;
            }
        }
        if(flag)break;
    }
    if(!flag)return 0;
    return 1e9;
}
int dfs(int num){
    int rst;
    if(num==1)rst=-1e9;
    else rst=1e9;
    int flag=check();
    if(flag!=1e9)return flag;
    rep(i,1,3){
        rep(j,1,3){
            if(mp[i][j]==0){
                mp[i][j]=num;
                if(num==1)rst=max(rst,dfs(2));
                else rst=min(rst,dfs(1));
                mp[i][j]=0;
            }
        }
    }
    return rst;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        rep(i,1,3){
            rep(j,1,3){
                scanf("%d",&mp[i][j]);
            }
        }
        int ans=dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}