CCF-CSP题解 201803-4 棋局评估

求当前井字棋局的得分。

用dfs虚构一下搜索树，每个节点对应一个不同的棋局。

每个节点有一个situation()情况评估，若胜负已定，则对应该棋局的评分；否则为0，表示胜负未定或平局。

每个节点还有一个得分用于return，如果situation()值不为0，胜负已定，则节点不再向下拓展，得分即为situation()值；否则若棋盘已满为平局，得分为0，若棋盘未满胜负未定，节点向下拓展，得分需要根据子节点的得分及当前下棋人cur确定。

出题人有一句“当棋盘被填满的时候，游戏结束，双方平手”。Absolutely wrong！棋盘填满不一定平手，一定是先要situation()为0再判断棋盘满不满，以确定是否平手。

#include <bits/stdc++.h>

using namespace std;

struct tNode
{
    int chess[9];
    tNode()
    {
        memset(chess, 0, sizeof(chess));
    }
    tNode(tNode *y)
    {
        for (int i = 0; i <= 8; i++)
            chess[i] = y->chess[i];
    }
    int remain()
    {
        int ret = 0;
        for (int i = 0; i <= 8; i++)
        {
            if (chess[i] == 0)
                ret ++;
        }
        return ret;
    }
    int situation()
    {
        if (chess[0] == chess[3] && chess[3] == chess[6] && chess[0] != 0)
        {
            if (chess[0] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[1] == chess[4] && chess[4] == chess[7] && chess[1] != 0)
        {
            if (chess[1] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[2] == chess[5] && chess[5] == chess[8] && chess[2] != 0)
        {
            if (chess[2] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[0] == chess[1] && chess[1] == chess[2] && chess[0] != 0)
        {
            if (chess[0] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[3] == chess[4] && chess[4] == chess[5] && chess[3] != 0)
        {
            if (chess[3] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[6] == chess[7] && chess[7] == chess[8] && chess[6] != 0)
        {
            if (chess[6] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[0] == chess[4] && chess[4] == chess[8] && chess[0] != 0)
        {
            if (chess[0] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        if (chess[2] == chess[4] && chess[4] == chess[6] && chess[2] != 0)
        {
            if (chess[2] == 1)
                return 1 + remain();
            else
                return - (1 + remain());
        }
        return 0;
    }
};

int dfs(tNode *x, int cur)
{
    int sit = x->situation();
    if (sit != 0)
        return sit;
    if (x->remain() == 0)
        return 0;
    int mmax = -20, mmin = 20;
    for (int i = 0; i <= 8; i++)
    {
        if (x->chess[i] == 0)
        {
            tNode *xx = new tNode(x);
            xx->chess[i] = cur;
            int temp = dfs(xx, cur % 2 + 1);
            mmax = max(mmax, temp);
            mmin = min(mmin, temp);
        }
    }
    if (cur == 1)
        return mmax;
    else
        return mmin;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        tNode *st = new tNode();
        for (int i = 0; i <= 8; i++)
            scanf("%d", &st->chess[i]);

        printf("%d\n", dfs(st, 1));
    }
    return 0;
}