hdu 5724(博弈论+状态压缩)
来源:互联网 发布:pdo php extension 编辑:程序博客网 时间:2024/05/18 01:22
Chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1044 Accepted Submission(s): 451
Problem Description
Alice and Bob are playing a special chess game on an n × 20 chessboard. There are several chesses on the chessboard. They can move one chess in one turn. If there are no other chesses on the right adjacent block of the moved chess, move the chess to its right adjacent block. Otherwise, skip over these chesses and move to the right adjacent block of them. Two chesses can’t be placed at one block and no chess can be placed out of the chessboard. When someone can’t move any chess during his/her turn, he/she will lose the game. Alice always take the first turn. Both Alice and Bob will play the game with the best strategy. Alice wants to know if she can win the game.
Input
Multiple test cases.
The first line contains an integerT(T≤100) , indicates the number of test cases.
For each test case, the first line contains a single integern(n≤1000) , the number of lines of chessboard.
Thenn lines, the first integer of ith line is m(m≤20) , indicates the number of chesses on the ith line of the chessboard. Then m integers pj(1≤pj≤20) followed, the position of each chess.
The first line contains an integer
For each test case, the first line contains a single integer
Then
Output
For each test case, output one line of “YES” if Alice can win the game, “NO” otherwise.
Sample Input
212 19 2021 191 18
Sample Output
NOYES
Author
HIT
之前没学过博弈论,所以参照别人的代码改的。尽管还有些不太懂,但多做一些题应该就能理解了吧
#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=(1<<20)+5;int sg[maxn];int SG(int state){ if(sg[state]!=-1) return sg[state]; bool vis[22]; memset(vis,false,sizeof(vis)); for(int i=0;i<20;i++) { if(!(state&(1<<i))) continue; for(int j=i+1;j<20;j++) { if(state&(1<<j)) continue; vis[SG(state^(1<<i)^(1<<j))]=true; break; } } for(int i=0;i<22;i++) { if(!vis[i]) { sg[state]=i;break; } } return sg[state];}int main(){ memset(sg,-1,sizeof(sg)); for(int i=0;i<(1<<20);i++) SG(i); int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); int ans=0; for(int i=0;i<n;i++) { int m,tmp=0,tmp1; scanf("%d",&m); for(int j=0;j<m;j++) { scanf("%d",&tmp1); tmp |= (1<<(tmp1-1)); } ans^=sg[tmp]; } if(ans==0) printf("NO\n"); else printf("YES\n"); } return 0;}
0 0
- hdu 5724(博弈论+状态压缩)
- HDU 5724 Chess(博弈论)
- hdu 4739(状态压缩)
- Hdu 1429(状态压缩)
- hdu 3006(状态压缩)
- HDU 5724 Chess(状态压缩+组合博弈)
- hdu 5724 Chess (sg函数 + 状态压缩)
- POJ 1143 (记忆化搜索,状态压缩,博弈论记忆化搜索实现)
- hdu 2167 Pebbles(状态压缩)
- hdu 4284(状态压缩)&& poj 3311 &&,,,
- HDU 3001--Travelling(状态压缩+tsp)
- hdu - 4628 - Pieces(状态压缩dp)
- HDU 4628 Pieces(状态压缩dp)
- hdu 1185(状态压缩dp)
- hdu 4628(状态压缩dp)
- hdu 2167 Pebbles(状态压缩DP)
- hdu 3001(状态压缩dp)
- hdu 4628 Pieces(状态压缩DP)
- VS2015 +EF6 连接MYSQL数据库生成实体
- Android studio基本使用(1)
- 装饰设计模式的演示
- Spark学习使用笔记 - Scala篇(1)
- Android_百度地图,点聚合功能,点击事件&&设置聚合数字的背景颜色
- hdu 5724(博弈论+状态压缩)
- 计算机图形学(三)二维几何变换
- Python小技巧
- word2vec前世今生
- shell删除创建时间大于10天的所有文件
- Android Studio新功能解析,你真的了解Instant Run吗?
- [DP46题] HDU 1864 最大报销额
- 今天看看Java反射
- The Text Splitting CodeForces 612A