组合博弈学习笔记

组合博弈学习笔记

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说在前边

1. 下面的博弈题目基本就是sg函数，搜必败必胜态，找规律，推策略。。。没有对理论进行深入了解。

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HDU1527

搜索时发现，必败态的数对貌似有规律，首先他们的大小两个数的差值是逐个增加的。然后，差分打表，发现差值为1或者2.实在找不到规律了，OEIS了一发，是个黄金分割，学到了。

//**黄金分割**
//**a(n) = floor(n*phi), where phi = (1+sqrt(5))/2
//**1,3,4,6,8,9,11,12,14,16,17,19,21,22,24,25,27,29,30,32,33,35,37
//**特点: 在i附近，貌似有规律地浮动
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef long long ll;
using namespace std;
bool f[1101][1011],vis[1112][1101];
int dfs(int x,int y) {if(x==0&&y==0) return 0;if(vis[x][y]) return f[x][y];vis[x][y]=vis[y][x]=1;rep(i,1,x) if(!dfs(x-i,y)) return f[x][y]=f[y][x]=1;rep(i,1,y) if(!dfs(x,y-i)) return f[x][y]=f[y][x]=1;rep(i,1,min(x,y)) if(!dfs(x-i,y-i)) return f[x][y]=f[y][x]=1;return f[x][y]=f[y][x]=0;
}
vector<int> v;
int main() {
//**
//    rep(i,1,550) rep(j,1,i) {
//        if(!dfs(i,j))printf("(%d,%d) = %d\n",i,j,dfs(i,j)),v.pb(j);//**第i对数之间差i
//    }
//    cout << (sqrt(5)+1)/2.0 << endl;
//    rep(i,0,(int)v.size()-1) {
//        printf("%d, ",v[i]);
//    }puts("");
//**ll a,b;while(scanf("%lld%lld",&a,&b) != EOF) {if(a>b) swap(a,b);ll t = b-a;ll ta = floor((sqrt(5.0)+1.0)*t/2.0);if(ta == a) puts("0");else puts("1");}return 0;
}

HDU1760

直接搜索即可，hash去重直接用map，hash二进制压位。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
typedef unsigned long long ull;
const int N = 55;
const int mod = 1000007;
using namespace std;
int n,m;
map<ull,bool> s;
char mp[N][N];
bool ck(int x,int y) {if(x>=1&&x+1<=n&&y>=1&&y+1<=m) {if(mp[x][y]=='0'&&mp[x][y+1]=='0'&&mp[x+1][y]=='0'&&mp[x+1][y+1]=='0')return 1;return 0;}return 0;
}
bool ack() {rep(i,1,n-1)rep(j,1,m-1)if(ck(i,j))return 1;return 0;
}
void put(int x,int y) {mp[x][y] = mp[x+1][y] = mp[x][y+1] = mp[x+1][y+1] = '1';
}
void del(int x,int y) {mp[x][y] = mp[x+1][y] = mp[x][y+1] = mp[x+1][y+1] = '0';
}
ull HS() { //hash 好好写ull hs = 0;rep(i,1,n) rep(j,1,m)if(mp[i][j]=='0')hs |= (1LL<<((i-1LL)*m+j-1LL));return hs;
}
int dfs() {ull hs = HS();if(s.find(hs)!=s.end()) return s[hs];if(!ack()) return s[hs]=0;int ans = 0;rep(i,1,n-1)rep(j,1,m-1) if(ck(i,j)){put(i,j);if(!dfs()) ans |= 1;del(i,j);}return s[hs]=ans;
}
int main() {while(scanf("%d%d",&n,&m) != EOF) {rep(i,1,n) scanf(" %s",mp[i]+1);s.clear();if(dfs())puts("Yes");else puts("No");}return 0;
}

HDU1847

直接预处理sg函数即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
typedef long long ll;
const int N = 1010;
using namespace std;
int n, sg[N];
bool vis[N];
void init() {sg[0] = 0;rep(s,1,1000) {memset(vis,0,sizeof(vis));rep(i,0,10)if((s>=(1<<i)))vis[sg[s - (1<<i)]] = 1;rep(i,0,1000) if(!vis[i]) {sg[s] = i;break;}}
}
int main() {init();while(scanf("%d",&n)!=EOF) {if(sg[n] == 0) puts("Cici");else puts("Kiki");}return 0;
}

HDU2516

记忆化搜索必败态，必胜态，发现满足斐波那契数列。之前的dp写法我自己都没看懂。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef long long ll;
using namespace std;//***
int dfs(int n,int x) {if(n==0)return 0;rep(i,1,min(2*x,n)) if(!dfs(n-i,i)) return 1;return 0;
}
int cal(int n) {rep(i,1,n-1) if(!dfs(n-i,i)) return 1;return 0;
}
void _init(int n) {rep(i,2,n) if(!cal(i)) printf("%d ",i);puts("");
}
//***vector<ll> v;
void init() {ll f0 = 1, f1 = 1, f2 = 2;while(f2<(1LL<<32LL)) {v.pb(f2);f0 = f1; f1 = f2; f2 = f1 + f0;}
//    for(auto x: v)printf("%lld ",x);puts("");
}
int main() {//**//_init(50);//**init();ll n;while(scanf("%lld",&n),n) {if(*lower_bound(v.begin(),v.end(),n)==n)puts("Second win");else puts("First win");}return 0;
}

HDU3951

直接暴力求一根链情况下的sg函数，然后发现长度大于1必胜。环的情况就很容易了。还有一种思路是对称的取硬币可以证明后手必胜。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
typedef long long ll;
const int N = 1e3 + 7;
using namespace std;
int sg[N],vis[N];
void init(int K) {sg[0] = 0;rep(s,1,100) {memset(vis,0,sizeof(vis));rep(k,1,K) rep(i,1,s) {if(s-(i+k-1)>=0) {vis[sg[i-1]^sg[s-(i+k-1)]]=s;}}rep(i,0,100) if(vis[i]!=s){sg[s]=i;break;}}rep(s,1,100)printf("%d ",sg[s]);puts("");
}
int main() {//**//rep(i,1,10)init(i);//对于一根链//当k=1时，奇数必胜//当k大于1时，大于0就必胜//**int T,C=0;scanf("%d",&T);while(T--) {ll n; int K;scanf("%lld%d",&n,&K);printf("Case %d: ",++C);if(K==1) {if((n-1)%2==1)puts("second");else puts("first");}else {if(n<=K) puts("first");else puts("second");}}return 0;
}

HDU4559

将格子拆为 2×i 的格子与 1×1 的格子的组合，把他们分别处理sg值然后异或起来。对于 1×1 的格子，sg值就是1，对于2*i的格子，可以直接求sg值，具体写法见代码。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef long long ll;
const int N = 4750;
using namespace std;
int T,C=0,n,m,sg[N],vis[N],num[N];
void init() {sg[0] = sg[1] = 0;rep(s,2,4747) {rep(i,1,s) {int l = i-1, r = s-i;vis[sg[l]^1^sg[r]]=s; // 1*1if(i+1<=s) {l=i-1; r=s-(i+1);vis[sg[l]^sg[r]]=s; // 2*2}}rep(i,0,4747) if(vis[i]!=s){sg[s] = i; break;}}
}
int solve() {int t0, t1, ans;t0=t1=ans=0;rep(i,1,n) {if(num[i] == 1) t1^=1;if(num[i] == 0) ++t0;else {ans^=sg[t0];t0 = 0;}}(ans^=sg[t0])^=t1;return ans;
}
int main() {init();scanf("%d",&T);while(T--) {scanf("%d%d",&n,&m);rep(i,0,n) num[i]=vis[i]=0;rep(i,1,m) {int x,y;scanf("%d%d",&x,&y);++num[y];}printf("Case %d: ",++C);if(solve()) puts("Alice");else puts("Bob");}return 0;
}

HDU4642

每次反转一个位置右下角一定会翻转一次，如果Alice将它反转为0，则Bob一定会将它反转为1，则最后一定是Alice赢。反之同理。所以直接判最后一位是0还是1即可

#include <iostream>
#include <map>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
typedef long long ll;
using namespace std;
int n,m;
int main() {int T;scanf("%d",&T);while(T--) {scanf("%d%d",&n,&m);int x;rep(i,1,n)rep(j,1,m) scanf("%d",&x);if(x==1) puts("Alice");else puts("Bob");}return 0;
}

HDU5963

与上一题思路基本一致，子节点的修改都会对根节点上的对应边进行反转，所以直接判那条边是否为1即可，多条边，直接判断奇偶。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define MP make_pair
#define PII pair<int,int>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
map<PII,int> M;
int n,m,d[44444];
int main() {int T;scanf("%d",&T);while(T--) {scanf("%d%d",&n,&m);rep(i,1,n) d[i]=0; M.clear();rep(i,1,n-1) {int x,y,z;scanf("%d%d%d",&x,&y,&z);d[x]+=z;d[y]+=z;if(x > y) swap(x,y);M[MP(x,y)] = z;}rep(i,1,m) {int opt,x,y,z;scanf("%d",&opt);if(opt) {scanf("%d%d%d",&x,&y,&z);if(x>y)swap(x,y);int t = M[MP(x,y)];d[x]+=z; d[y]+=z;d[x]-=t; d[y]-=t;M[MP(x,y)] = z;}else {int x;scanf("%d",&x);if(d[x]%2)puts("Girls win!");else puts("Boys win!");}}}return 0;
}