题目链接
题意：井字棋，现在放了某些棋子。连成线的时候得分为(空格子数+1)（B赢*-1）问当前棋局中，如果Alice和Bob都按最优策略下棋，最终得分。
思路：典型的min-max对抗搜索，A选取分数最高的一种走法，B选取分数最低的一种走法。
注意的是dfs返回值的初始化问题，min搜索初始值不一定小于0， max初始值不一定大于0（因为这个愚蠢的错了好几回）。
代码：
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| #include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int graph[5][5];
int judge()
{
for(int i = 0; i < 3; i++)if(graph[0][i] != 0 && graph[0][i] == graph[1][i] && graph[2][i] == graph[0][i])return graph[0][i];
for(int i = 0; i < 3; i++)if(graph[i][0] != 0 && graph[i][0] == graph[i][1] && graph[i][2] == graph[i][0])return graph[i][0];
if(graph[0][0] != 0 && graph[0][0] == graph[1][1] && graph[0][0] == graph[2][2])return graph[0][0];
if(graph[2][0] != 0 && graph[2][0] == graph[1][1] && graph[2][0] == graph[0][2])return graph[2][0];
return 0;
}
int getdonow()
{
int cnt = 0;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if(graph[i][j] == 0)cnt++;
}
}
return cnt;
}
int dfs(int donow)
{
int person = donow % 2;
if(person == 0)person = 2;
if(judge() != 0)return person == 2? donow+1 : -donow-1;
if(donow == 0)return 0;
int retmax = -100, retmin = 100;
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
if(graph[i][j] == 0){
graph[i][j] = person;
if(person == 1)retmax = max(retmax, dfs(donow-1));
else retmin = min(retmin, dfs(donow-1));
graph[i][j] = 0;
}
}
}
if(person == 1)return retmax;
else return retmin;
}
int main()
{
int T;
int flag;
int donow;
scanf("%d", &T);
while(T--)
{
memset(graph, 0, sizeof(graph));
for(int i = 0; i < 3; i++)
{
scanf("%d%d%d", &graph[i][0], &graph[i][1], &graph[i][2]);
}
donow = getdonow();
int ans = dfs(donow);
if(ans > 0 && ans % 2 == 0)ans++;
else if(ans < 0 && ans % 2 == 1)ans --;
printf("%d\n", ans);
}
return 0;
}
|