5248: [2018多省省队联测]一双木棋

分析：

　　极大极小搜索！记忆化一下，所有的状态是阶梯状的！然后缩成一个LL。

　　好像状压dp也行。

　　考试的时候。。。啥都不会。。。不会极大极小搜索，不会dp，不会这道题！

代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 
 5 inline int read() {
 6     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
 7     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
 8 }
 9 
10 const int N = 20;
11 const LL base = 11;
12 int a[N][N],b[N][N],sta[N],n,m;
13 map<LL,int> p;
14 
15 inline LL getHash() {
16     LL ret = 0;
17     for (int i=1; i<=n; ++i) ret = ret * base + sta[i];
18     return ret;
19 }
20 inline void getState(LL S) {
21     for (int i=n; i>=1; --i) sta[i] = S % base,S /= base;
22 }
23 int Minimax(int player,LL S) {
24     if (p.find(S) != p.end()) return p[S];
25     getState(S);
26     int res = player ? 1e9 : -1e9;
27     for (int i=1; i<=n; ++i) {
28         if (sta[i-1] > sta[i]) {
29             ++ sta[i];
30             LL T = getHash();
31             if (!player) res = max(res,Minimax(player^1,T) + a[i][sta[i]]) ;
32             else res = min(res,Minimax(player^1,T) - b[i][sta[i]]) ;
33             -- sta[i];
34         }
35     }
36     p[S] = res;
37     return res;
38 }
39 
40 int main() {
41     n = read(),m = read();
42     for (int i=1; i<=n; ++i) 
43         for (int j=1; j<=m; ++j) a[i][j] = read();
44     for (int i=1; i<=n; ++i) 
45         for (int j=1; j<=m; ++j) b[i][j] = read();
46     
47     for (int i=1; i<=n; ++i) sta[i] = m;
48     p[getHash()] = 0;
49     
50     sta[0] = m;
51     
52     cout << Minimax(0,0);
53     
54     return 0;
55 }