算法筆記
參考資料：https://wenku.baidu.com/view/25540742a8956bec0975e3a8.html
sg函數大神詳解：http://blog.csdn.net/luomingjun12315/article/details/45555495
sg[i]定義，從i走一步能到達的j的sg[j]以外的最小值，那么從sg函數值為x的狀態出發，我們能轉移到sg值為0,1,...,x-1的狀態
對於某個人來說，0是他的必敗態，sg[0] = 0
我們從這個狀態出發，用dp求sg函數的值
sg[n] = 0，表示必敗，否則， 表示必勝
如果sg[n] > 0，說明肯定能轉移到必敗態，則必勝
如果sg[n] = 0, 說明無論怎么轉移都是必勝態，則必敗
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]++; } }
例題：http://www.cnblogs.com/widsom/p/7171428.html
http://www.cnblogs.com/widsom/p/7170891.html
sg函數拓展：
反sg博弈：
先手必勝：（所有單一局面sg值都不超過1&&總局面sg值為0） || （存在一個單一局面sg值超過1&&總局面sg值不為0）
否則后手必勝。
代碼：
#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; }
樹上刪邊博弈：
定理：葉子節點的sg值為0，其他節點u的sg[u]值等於它兒子v的(sg[v]+1)的亦或和。
圖上刪邊博弈：
將偶環縮成點，奇環縮成一個點加一條邊，就可以轉換成樹上刪邊博弈了。
具體證明看最上面的鏈接。
思路：樹上刪邊博弈
代碼：
#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; }
思路：tarjan縮邊雙轉換成樹上刪邊博弈
代碼：
#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標記環上的點是否被刪掉 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; }
思路：
圓掃描線+樹上刪邊博弈
代碼：
#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; }
思路：
出題人真是個機靈鬼，將反-sg和樹上刪邊結合起來，大概是看了論文后才出的題（霧
代碼：
#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; }