【题解】POJ-2425 A Chess Game

A Chess Game(POJ-2425)

题面

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!

输入

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.

输出

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.

样例输入

1
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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

样例输出

1
2
3
4
5
WIN
WIN
WIN
LOSE
WIN

提示

Huge input,scanf is recommended.

思路

给定一个 n 个节点的有向无环图,再给出 m 个棋子,每一个棋子位于一个点上且互相不影响(即一个点上可以有多个棋子),每回合可以选择一个棋子按照给出的图把它移向下一个点,即当前点与下一个点间有一条有向边,由当前点指向下一个点,无法进行操作则判输,问先手是否必胜。

一个棋子不能再移动当且仅当这个点的出度为0,因为每个棋子互不影响,所以分别求出每个棋子的SG函数,求一下异或和就好了。

代码

查看代码
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using namespace std;
const int mxn = 1e3 + 5;

vector<int> E[mxn];
int SG[mxn];

int getSg(int x){
if(SG[x]!=-1) return SG[x];
bool S[mxn] = {0};

for(int i=0; i<E[x].size(); i++){
S[getSg(E[x][i])] = 1;
}

int mex = 0;
while(S[mex]) mex++;
return SG[x]=mex;
}

int main(){
int n;
while(~scanf("%d", &n) && n)
{
memset(SG, -1, sizeof(SG));
for(int i=0; i<n; i++){
E[i].clear();
int m; scanf("%d", &m);
while(m--){
int x; scanf("%d", &x);
E[i].push_back(x);
}
}
int q;
while(~scanf("%d", &q) && q){
int nim = 0;
while(q--){
int x; scanf("%d", &x);
nim ^= getSg(x);
}
if(nim){
printf("WIN\n");
}else{
printf("LOSE\n");
}
}
}
}
_/_/_/_/_/ EOF _/_/_/_/_/