HDU1524-A Chess Game

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本文介绍了一个基于有向无环图的游戏，玩家轮流移动棋子，直至一方无法移动而告负。通过搜索+SG+nim异或操作策略来解决此问题，并提供了完整的C++代码实现。

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题目链接 A Chess Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2523 Accepted Submission(s): 1139

Problem Description

Let's design a new chess game. There are N positions to hold M chesses in this game. Multiple chesses can be located in the same position. The positions are constituted as a topological graph, i.e. there are directed edges connecting some positions, and no cycle exists. Two players you and I move chesses alternately. In each turn the player should move only one chess from the current position to one of its out-positions along an edge. The game does not end, until one of the players cannot move chess any more. If you cannot move any chess in your turn, you lose. Otherwise, if the misfortune falls on me... I will disturb the chesses and play it again.

Do you want to challenge me? Just write your program to show your qualification!

Input

Input contains multiple test cases. Each test case starts with a number N (1 <= N <= 1000) in one line. Then the following N lines describe the out-positions of each position. Each line starts with an integer Xi that is the number of out-positions for the position i. Then Xi integers following specify the out-positions. Positions are indexed from 0 to N-1. Then multiple queries follow. Each query occupies only one line. The line starts with a number M (1 <= M <= 10), and then come M integers, which are the initial positions of chesses. A line with number 0 ends the test case.

Output

There is one line for each query, which contains a string "WIN" or "LOSE". "WIN" means that the player taking the first turn can win the game according to a clever strategy; otherwise "LOSE" should be printed.

Sample Input

2 1 2

1 3

1 0

2 0 2

1 1

1 2

2 0 1

2 1 1

3 0 1 3

Sample Output

WIN

LOSE

WIN

Source

PKU Monthly

题目意思，给你一个有向无环图，两个人将棋子在上移动，如果谁不能移动那他就输了(移动到出度为零的点就不能再移动了)。

思路:搜索+sg+nim异或操作

代码:和这位大佬思路一样，但是我没能实现TAT,所以借鉴了一下大佬的。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector> 
using namespace std;
vector<int>vi[1010];
int g[1010];
int getSG(int x) {
    if(g[x] != -1) return g[x];//等于-1说明已经赋值过了，无需进行搜索 
    if(vi[x].size() == 0) return 0;//没有可以到达的点就直接返回 
    int vis[1010];
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < vi[x].size(); i++) {
        g[vi[x][i]] = getSG(vi[x][i]);
        vis[g[vi[x][i]]] = 1;
    }
    for(int i = 0;; i++){
    	if(!vis[i]) return i;
	}
}
int main() {
    int n,m,temp;
    while(~scanf("%d", &n)) {
        memset(g, -1, sizeof(g));
        for(int i = 0; i <n; i++){
        	vi[i].clear();
			scanf("%d", &m);
            for(int j = 1; j <= m; j++) {
                scanf("%d", &temp);
                vi[i].push_back(temp);
            }       	
        }
        while(scanf("%d", &m),m){
            int sum = 0;
            while(m--){
                scanf("%d", &temp);
                if(g[temp] == -1)
                    g[temp] = getSG(temp);
                sum^= g[temp];
            }
            if(sum == 0) puts("LOSE");
            else puts("WIN");
        }
    }
}