HDU 1524 A Chess Game(SG博弈)

A Chess Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1966    Accepted Submission(s): 884

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

42 1 201 301 02 0 2041 11 2002 0 12 1 13 0 1 30

 

Sample Output

WINWINWINLOSEWIN

 

Source

PKU Monthly

 

题意：有N个位置，其中存在拓扑关系，移动时必须遵守。最后移动者胜，问是否有必胜策略

AC代码，直接用Vector存储顶点与顶点的关系即可

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#define FIN freopen("input.txt", "r", stdin)using namespace std;const int MAXN = 1e3 + 5;int par[MAXN];int SG[MAXN];vector<int>G[MAXN];void init(int n){    memset(SG, -1, sizeof(SG));    for(int i = 0;i <= n; i ++) G[i].clear();}int getSG(int x){    int Hash[MAXN] = {0};    for(int i = 0;i < G[x].size();i ++){        if(SG[G[x][i]] == -1) SG[G[x][i]] = getSG(G[x][i]);        Hash[SG[G[x][i]]] = 1;    }    for(int i = 0;i < MAXN;i ++){        if(!Hash[i]){            return i;        }    }}int main(){    //FIN;    int N, n, x, m;    while(~scanf("%d", &N)){        init(N);        for(int i = 0;i < N;i ++){            scanf("%d", &n);            while(n --){                scanf("%d", &x);                G[i].push_back(x);            }        }        while(~scanf("%d", &m), m){            int sum = 0;            while(m --){                scanf("%d", &x);                if(SG[x] == -1) SG[x] = getSG(x);                sum ^= SG[x];            }            puts(sum ? "WIN" : "LOSE");        }    }    return 0;}

总结:

此刻起，才算真正的明白，SG函数真的就只是一个有向无环图，恩，感觉以后遇到这样的题基本无压力

但是有一个错误代码一直找不出原因，希望大牛可以帮忙看一看，与上面的vector实现方式基本是一样的

只是下面这份代码使用for循环来检测当前移动的位置是否为上一个位置的子节点。

#include <cstdio>#include <cstring>#include <algorithm>#define FIN freopen("input.txt", "r", stdin)using namespace std;const int MAXN = 2e3 + 5;int par[MAXN];int SG[MAXN];void init(){    for(int i = 0;i < MAXN;i ++) par[i] = -1;    memset(SG, -1, sizeof(SG));}int getSG(int x, int n){    int Hash[MAXN] = {0};    int i;    for(i = 1;i <= n;i ++){        if(x - i < 0) break;        if(par[x - i] != x) continue;        if(SG[x - i] == -1){            SG[x - i] = getSG(x - i, n);        }        Hash[SG[x - i]] = 1;    }    for(i = 1;i <= n;i ++){        if(x + i > n) break;        if(par[x + i] != x) continue;        if(SG[x + i] == -1){            SG[x + i] = getSG(x + i, n);        }        Hash[SG[x + i]] = 1;    }    for(i = 0;i < n ;i ++){        if(!Hash[i]){            return i;        }    }}int main(){    //FIN;    int N, n, x, m;    while(~scanf("%d", &N)){        init();        for(int i = 0;i < N;i ++){            scanf("%d", &n);            while(n --){                scanf("%d", &x);                par[x] = i;            }        }        while(~scanf("%d", &m), m){            int sum = 0;            while(m --){                scanf("%d", &x);                if(SG[x] == -1) SG[x] = getSG(x, N);                sum ^= SG[x];            }            puts(sum ? "WIN" : "LOSE");        }    }    return 0;}