POJ2311 Cutting Game(博弈论）

标签：quit   var   namespace   sig   problem   origin   内存限制   contains   nta   

总时间限制: 1000ms 内存限制: 65536kB
描述
Urej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those games can show his extraordinary wit. Recently Urej takes a great interest in a new game, and Erif Nezorf becomes the victim. To get away from suffering playing such a dull game, Erif Nezorf requests your help. The game uses a rectangular paper that consists of W*H grids. Two players cut the paper into two pieces of rectangular sections in turn. In each turn the player can cut either horizontally or vertically, keeping every grids unbroken. After N turns the paper will be broken into N+1 pieces, and in the later turn the players can choose any piece to cut. If one player cuts out a piece of paper with a single grid, he wins the game. If these two people are both quite clear, you should write a problem to tell whether the one who cut first can win or not.
输入
The input contains multiple test cases. Each test case contains only two integers W and H (2 <= W, H <= 200) in one line, which are the width and height of the original paper.
输出
For each test case, only one line should be printed. If the one who cut first can win the game, print "WIN", otherwise, print "LOSE".
样例输入
2 2
3 2
4 2
样例输出
LOSE
LOSE
WIN
来源
POJ Monthly,CHEN Shixi(xreborner)

题解：
用sg函数做，还套了个记忆化
切成两半以后，返回的sg值就是两个数的异或。如果有两个必胜，用掉一个，别人再用一个，你就是两个必败了
注意点：为什么要从2开始枚举？
因为如果是1了，先手必胜(非（1,1），可是我的初始化没有设置这种情况，就会出现-1

#include <bits/stdc++.h>
#define int long long
using namespace std;
int sg[230][230], x, y, tong[1200];
bool vis[230][230];
int find(int x, int y) {
    if(sg[x][y]!=-1) return sg[x][y];
    memset(tong, 0, sizeof tong);
    for (int i = 2; i <= x-i; i++)
        tong[(find(i, y)) ^ (find(x - i, y))]=1;
    for (int i = 2; i <= y-i; i++)
        tong[(find(x, i)) ^ (find(x, y - i))]=1;
    for (int i = 0; i < 1200; i++)
        if (!tong[i]) {
            sg[x][y] = i;
            return i;
        }
}
signed main() {
    memset(sg,-1,sizeof sg);
    sg[2][2] = sg[2][3] = sg[3][2] = 0;
    while (scanf("%lld%lld", &x, &y) != EOF) {
        if (!find(x, y))
            puts("LOSE");
        else
            puts("WIN");
    }
    return 0;
}

POJ2311 Cutting Game(博弈论）

标签：quit   var   namespace   sig   problem   origin   内存限制   contains   nta   

原文地址：https://www.cnblogs.com/wky32768/p/10783126.html