POJ 2425 A Chess Game 博弈+有向无环图

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本文介绍了一个结合博弈论与有向无环图的游戏问题，通过设计算法判断玩家在特定规则下的胜负情况。文章详细阐述了解题思路及实现代码。

A Chess Game

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3821		Accepted: 1565

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

Sample Output

WIN
WIN
WIN
LOSE
WIN

Hint

Huge input,scanf is recommended.

Source

PKU Monthly,CHEN Shixi(xreborner)

博弈继续练习中。

这道题是博弈+有向无环图。

题意：一个有向无环图上有n个点（1<= n <= 1000），每个点上有一些有向边，连向其他的点。初始时在一些点上共放了m个棋子（1<= m <= 10）。两人进行博弈游戏，每个人可以任选一枚棋子将其移动到任意可一步到达的位置，无法移动棋子者输。问先手赢还是输。

解题思路：很明显这个题是组合游戏，将每个棋子初始位置的sg值求出来之后再进行nim亦或就行了。一开始我以为要将所有点的sg值用dp预处理出来，但是题中说棋子数不超过10，所以可以输入的同时用dfs计算sg值，也不会超时。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <numeric>
#include <iomanip>
#include <climits>
#include <new>
#include <utility>
#include <iterator>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>

using namespace std;

int path[1005][1005];
int sg[1005];
int a[15];

int dfs(int v)
{
    if (sg[v] != -1)
        return sg[v];
    bool vis[1005];
    memset(vis,false, sizeof(vis));
    int tmp;
    for (int i = 1; i <= path[v][0]; ++i) {
        tmp = dfs(path[v][i]);
        vis[tmp] = true;
    }
    int i = 0;
    while (vis[i] == true)  i++;
    sg[v] = i;
    return sg[v];

}

int main()
{
    int n;
    while (scanf("%d", &n) != EOF) {
        memset(path,-1,sizeof(-1));
        memset(sg,-1,sizeof(sg));
        for (int i = 0; i < n; ++i) {
            scanf("%d", &path[i][0]);
            if(path[i][0] == 0)
                sg[i] = 0;
            for (int j = 1;  j <= path[i][0]; ++j) {
                scanf("%d", &path[i][j]);
            }
        }
        int num;
        int ans = 0,a;
        while (scanf("%d",&num) && num) {
            ans = 0;
            for(int i = 0; i < num; ++i) {
                scanf("%d", &a);
                ans ^= dfs(a);
            }
            if(ans == 0)    printf("LOSE\n");
            else printf("WIN\n");
        }
    }
    return 0;
}