HDU 6139 Galaxy at War(2017 Multi-University Training Contest 8)
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题目链接:Galaxy at War
题意:在一个给定的
题解:这个游戏是一个简单的阶梯博弈,因为在
#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>using namespace std;const int N = 500005;pair<int,int> med[N],pol[N];struct Ct{ int x,y,w; Ct(){} Ct(int x,int y,int w):x(x),y(y),w(w){}};vector<Ct>c;int main(){ int T,n,m,k,x,y,w,t,s,mx; scanf("%d",&T); while(T--){ mx=0; scanf("%d %d %d",&n,&m,&k); c.clear(); for(int i=1;i<=k;i++){ scanf("%d %d %d",&x,&y,&w); c.push_back(Ct(x,y,w)); } scanf("%d",&t); for(int i=1;i<=t;i++){ scanf("%d %d",&med[i].first,&med[i].second); if(med[i].first==n) mx=max(mx,med[i].second); } scanf("%d",&s); for(int i=1;i<=s;i++) scanf("%d %d",&pol[i].first,&pol[i].second); int ans=0; for(int i=0;i<k;i++){ if(c[i].x<n&&(abs(c[i].x-n+1)+abs(c[i].y-m))%2==0) ans^=c[i].w; else if(c[i].x==n&&c[i].y<=mx&&abs(c[i].y-mx)%2==0) ans^=c[i].w; } puts(ans?"win":"lose"); } return 0;}
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