大家好，又见面了，我是全栈君。
算法笔记
参考资料：https://wenku.baidu.com/view/25540742a8956bec0975e3a8.html
sg函数大神详解：http://blog.csdn.net/luomingjun12315/article/details/45555495
sg[i]定义，从i走一步能到达的j的sg[j]以外的最小值，那么从sg函数值为x的状态出发，我们能转移到sg值为0,1,…,x-1的状态
对于某个人来说，0是他的必败态，sg[0] = 0
我们从这个状态出发，用dp求sg函数的值
sg[n] = 0，表示必败，否则， 表示必胜
如果sg[n] > 0，说明肯定能转移到必败态，则必胜
如果sg[n] = 0, 说明无论怎么转移都是必胜态，则必败
模板：
int f[N],SG[N]; bool S[M]; void getSG(int n) { memset(SG,0,sizeof(SG)); for(int i=1;i<=n;i++) { memset(S,false,sizeof(S)); for(int j=1;f[j]<=i&&j<M;j++) { S[SG[i-f[j]]]=true; } while(S[SG[i]]) SG[i]++; } }
例题：http://www.cnblogs.com/widsom/p/7171428.html
http://www.cnblogs.com/widsom/p/7170891.html
sg函数拓展：
反sg博弈：
先手必胜：（所有单一局面sg值都不超过1&&总局面sg值为0） || （存在一个单一局面sg值超过1&&总局面sg值不为0）
否则后手必胜。
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; const int N = 55, M = 5e3 + 5; int a[N], sg[M], T, n; int main() { for (int i = 0; i < M; ++i) sg[i] = i; scanf("%d", &T); while(T--) { scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", &a[i]); int cnt = 0, s = 0; for (int i = 1; i <= n; ++i) { if(sg[a[i]] > 1) ++cnt; s ^= sg[a[i]]; } if((!cnt && !s) || (cnt && s)) printf("John\n"); else printf("Brother\n"); } return 0; }
View Code
树上删边博弈：
定理：叶子节点的sg值为0，其他节点u的sg[u]值等于它儿子v的(sg[v]+1)的亦或和。
图上删边博弈：
将偶环缩成点，奇环缩成一个点加一条边，就可以转换成树上删边博弈了。
具体证明看最上面的链接。
思路：树上删边博弈
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; const int N = 1e5 + 5; vector<int> g[N]; int T, n, u, v; int sg(int u, int o) { int res = 0; for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v != o) res ^= sg(v, u) + 1; } return res; } int main() { scanf("%d", &T); while(T--) { scanf("%d", &n); for (int i = 1; i < n; ++i) scanf("%d %d", &u, &v), g[u].pb(v), g[v].pb(u); if(sg(1, 1)) printf("Alice\n"); else printf("Bob\n"); for (int i = 1; i <= n; ++i) g[i].clear(); } return 0; }
View Code
思路：tarjan缩边双转换成树上删边博弈
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<cstdio> #include<iostream> #include<cstring> #include<vector> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; const int N = 105; vector<int> g[N]; int t, n, m, u, v; int stk[N], sg[N], low[N], dfn[N], cnt = 0, top = 0; bool vis[N], vv[N];//vv标记环上的点是否被删掉 void tarjan(int u, int o) { dfn[u] = low[u] = ++cnt; stk[++top] = u; vv[u] = vis[u] = true; for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v == o) continue; if(!dfn[v]) tarjan(v, u), low[u] = min(low[u], low[v]); else if(vis[v]) low[u] = min(low[u], dfn[v]); } if(low[u] == dfn[u]) { int c = 0; while(stk[top] != u) { vv[stk[top]] = false; vis[stk[top--]] = false; ++c; } vis[stk[top--]] = false; ++c; if(c > 1 && c%2) sg[u] ^= 1; } for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v == o) continue; if(vv[v]) sg[u] ^= sg[v]+1; } } int main() { while(~scanf("%d", &t)) { int s = 0; while(t--) { scanf("%d %d", &n, &m); for (int i = 0; i < m; ++i) { scanf("%d %d", &u, &v); g[u].pb(v); g[v].pb(u); } tarjan(1, 1); s ^= sg[1]; for (int i = 1; i <= n; ++i) low[i] = dfn[i] = sg[i] = vis[i] = vv[i] = 0; cnt = top = 0; for (int i = 1; i <= n; ++i) g[i].clear(); } if(s) printf("Sally\n"); else printf("Harry\n"); } return 0; }
View Code
思路：
圆扫描线+树上删边博弈
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; const int N = 2e4 + 5; int nowx; struct circle { int x, y, r; }p[N]; double Y(int id, int ty) { if(ty == 0) return p[id].y - sqrt(p[id].r*1.0*p[id].r - (nowx-p[id].x)*1.0*(nowx-p[id].x)); else return p[id].y + sqrt(p[id].r*1.0*p[id].r - (nowx-p[id].x)*1.0*(nowx-p[id].x)); } struct node { int id, ty; bool operator < (const node &rhs) const { if(id == rhs.id) return ty < rhs.ty; else return Y(id, ty) < Y(rhs.id, rhs.ty); } }; set<node> s; vector<int> g[N]; int T, n, dp[N], fa[N], sg[N]; piii t[N*2]; void dfs(int u, int o) { sg[u] = 0; for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v != o) { dfs(v, u); sg[u] ^= sg[v] + 1; } } } int main() { p[0].x = p[0].y = 0; p[0].r = 100000; s.insert({ 0, 0}); s.insert({ 0, 1}); scanf("%d", &T); while(T--) { scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].r); for (int i = 1; i <= n; ++i) { t[i].fi.fi = p[i].x - p[i].r; t[i].fi.se = 0; t[i].se = i; t[n+i].fi.fi = p[i].x + p[i].r; t[n+i].fi.se = 1; t[n+i].se = i; } sort(t+1, t+1+2*n); for (int i = 1; i <= 2*n; ++i) { nowx = t[i].fi.fi; int id = t[i].se; node tmp = {id, 1}; if(t[i].fi.se == 0) { auto l = s.lower_bound(tmp); --l; auto r = s.upper_bound(tmp); if((*l).id == (*r).id) { dp[id] = dp[(*l).id] + 1; fa[id] = (*l).id; } else if(dp[(*l).id] >= dp[(*r).id]) { dp[id] = dp[(*l).id]; fa[id] = fa[(*l).id]; } else { dp[id] = dp[(*r).id]; fa[id] = fa[(*r).id]; } g[fa[id]].pb(id); s.insert({id, 1}); s.insert({id, 0}); } else { s.erase({id, 1}); s.erase({id, 0}); } } dfs(0, 0); if(sg[0]) printf("Alice\n"); else printf("Bob\n"); for (int i = 0; i <= n; ++i) g[i].clear(), sg[i] = fa[i] = dp[i] = 0; } return 0; }
View Code
思路：
出题人真是个机灵鬼，将反-sg和树上删边结合起来，大概是看了论文后才出的题（雾
代码：
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; const int N = 105; vector<int> g[N]; int t, n, u, v; int dfs(int u, int o) { int sg = 0; for (int i = 0; i < g[u].size(); ++i) { int v = g[u][i]; if(v != o) sg ^= dfs(v, u) + 1; } return sg; } int main() { while(~scanf("%d", &t)) { int cnt = 0, s = 0; while(t--) { scanf("%d", &n); for (int i = 1; i < n; ++i) scanf("%d %d", &u, &v), g[u].pb(v), g[v].pb(u); int sg = dfs(1, 1); s ^= sg; if(sg > 1) ++cnt; for (int i = 0; i <= n; ++i) g[i].clear(); } if((cnt && s) || (!cnt && !s)) printf("PP\n"); else printf("QQ\n"); } return 0; }
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转载于:https://www.cnblogs.com/widsom/p/7172386.html
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