poj 2425 A Chess Game（SG函数）

 

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　A Chess Game

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3551		Accepted: 1440

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

PKU Monthly,CHEN Shixi(xreborner)

 

【思路】

       SG函数。

       把每个询问视作一组游戏，ans为每组游戏sg函数的xor值。

       sg(x)=mex(S)，其中mex(S)表示取后继集合所没有出现的sg值中最小的一个。

      

　　做这道题学到了一个有意思的经验：因为vis每一层dfs都会独立用到，所以如果放在dfs外面声明为全局变量的话vis就会被接下来的dfs所覆盖。      

 

【代码】

　　

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#define FOR(a,b,c) for(int a=(b);a<(c);a++)
using namespace std;

const int N = 1000+10;

int n,m,sg[N];
vector<int> G[N];

int dfs(int u) {
    if(sg[u]!=-1) return sg[u];
    int vis[N];
    memset(vis,0,sizeof(vis));
    FOR(i,0,G[u].size())
        vis[dfs(G[u][i])]=1;
    for(int i=0;;i++)
        if(!vis[i]) return sg[u]=i;
}
int main() {
    while(scanf("%d",&n)==1) {
        FOR(i,0,n) G[i].clear();
        int v;
        FOR(i,0,n) {
            scanf("%d",&m);
            FOR(j,0,m) {
                scanf("%d",&v);
                G[i].push_back(v);
            }
        }
        memset(sg,-1,sizeof(sg));
        while(scanf("%d",&m) && m) {
            int ans=0,q;
            FOR(i,0,m)
                scanf("%d",&q) , ans^=dfs(q);
            if(ans) puts("WIN"); else puts("LOSE");
        }
    }
    return 0;
}