POJ2425 A Chess Game(博弈)

A Chess Game

Time Limit: 3000MS
Memory Limit: 65536K

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

4
2 1 2
0
1 3
0
1 0
2 0 2
0

4
1 1
1 2
0
0
2 0 1
2 1 1
3 0 1 3
0

Sample Output

WIN
WIN
WIN
LOSE
WIN

Hint

Huge input,scanf is recommended.

Source

POJ2425

题意

N个节点的有向无环图，节点上有M个棋子，每次可移动一个棋子，不能移动的一方输，给出图和初始棋子的位置，问先手胜负。

题解

多个棋子可视为多个游戏的组合，从题意可知出度为0的点为必败点，利用sg函数记忆化求出节点的sg值，对查询点的sg值互相取异或即可。AC代码如下

#include <cstdio>#include <cstring>using namespace std;int G[1002][1002];int Gnum[1002];int sg[1002];int N,M,x,shit;int ans(int p){    if(sg[p]!=-1)        return sg[p];    if(Gnum[p]==0)        return sg[p]=0;    bool vis[1002];    memset(vis,0,sizeof vis);    for(int i=0;i<Gnum[p];i++)        vis[ans(G[p][i])]=1;    for(int i=0;i<1002;i++)        if(vis[i]==0)            return sg[p]=i;}int main(){    while(scanf("%d",&N)!=EOF)    {        for(int i=0;i<N;i++)        {            Gnum[i]=0;            sg[i]=-1;            scanf("%d",&Gnum[i]);            for(int j=0;j<Gnum[i];j++)                scanf("%d",&G[i][j]);        }        while(scanf("%d",&M)&&M)        {            scanf("%d",&x);            shit=ans(x);            for(int i=1;i<M;i++)            {                scanf("%d",&x);                shit^=ans(x);            }            if(shit)                printf("WIN\n");            else                printf("LOSE\n");        }    }    return 0;}