題意：給一個井字棋的棋盤，對於已經贏的局面，得分是（棋盤上的空格子數+1）*（A為1，B為-1），給出現在的局面求最後的得分

思路：這個叫對抗搜索，每次換一個人搜一下，上次考我還在想下哪裏？結果答案是：搜索，隨便下

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #define LL long long
 6 #define debug(x) cout << "[" << x << "]" << endl
 7
 
 using namespace std;
 8 
 9 int a[3][3];
10 
11 bool row(int r, int p){
12     return a[r][0] == p && a[r][1] == p && a[r][2] == p;
13 }
14 
15 bool col(int c, int p){
16     return a[0][c] == p && a[1][c] == p && a[2][c] == p;
17 }
18 
19 int sum(){
20     int
 
 ans = 0;
21     for (int i = 0; i < 3; i++)
22         for (int j = 0; j < 3; j++)
23             if (!a[i][j]) ans++;
24     return ans;
25 }
26 
27 int win(int p){
28     int ans = 1;
29     bool ok = 0;
30     if (row(0, p) || row(1, p) || row(2, p)) ok = 1;
31     if (col(0, p) || col(1
 
, p) || col(2, p)) ok = 1;
32     if (a[0][0] == p && a[1][1] == p && a[2][2] == p) ok = 1;
33     if (a[0][2] == p && a[1][1] == p && a[2][0] == p) ok = 1;
34     if (!ok) return 0;
35     ans += sum();
36     return p == 1 ? ans : -ans;
37 }
38 
39 int dfs(int p){
40     if (!sum()) return 0;
41     int Max = -10, Min = 10;
42     for (int i = 0; i < 3; i++){
43         for (int j = 0; j < 3; j++){
44             if (a[i][j]) continue;
45             a[i][j] = p+1;
46             int w = win(p+1);
47             if (w){
48                 a[i][j] = 0;
49                 return w > 0 ? max(Max, w) : min(Min, w);
50             }
51             if (!p) Max = max(Max, dfs(1));
52             else Min = min(Min, dfs(0));
53             a[i][j] = 0;
54         }
55     }
56     return p ? Min : Max;
57 }
58 
59 int main(){
60     int t;
61     scanf("%d", &t);
62     while (t--){
63         for (int i = 0; i < 3; i++)
64             for (int j = 0; j < 3; j++)
65                 scanf("%d", &a[i][j]);
66         int x = win(1), y = win(2);
67         if (x) { printf("%d\n", x); continue; }
68         if (y) { printf("%d\n", y); continue; }
69         printf("%d\n", dfs(0));
70     }
71     return 0;
72 }

CCF 201803-4 棋局評估 （對抗搜索）