題目鏈接

BZOJ2437

題解

和JSOI2014很像
只不過這題動態刪點

如果我們把空位置看做\(X\)的話，就會發現我們走的路徑是一個\(OX\)交錯的路徑
然后將圖二分染色，當前點必勝，當且僅當當前點必須作為最大匹配的匹配點
移動相當于刪點，刪點的話只需打個標記即可
判斷當前點是不是必選，只需嘗試更改當前點匹配點的匹配點即可

#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt) #define REP(i,n) for (int i = 1; i <= (n); i++) #define mp(a,b) make_pair<int,int>(a,b) #define cls(s) memset(s,0,sizeof(s)) #define cp pair<int,int> #define LL long long int using namespace std; const int maxn = 1605,maxm = 100005,INF = 1000000000; inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag; } int h[maxn],ne = 1; struct EDGE{int to,nxt;}ed[maxm]; inline void build(int u,int v){ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;ed[++ne] = (EDGE){u,h[v]}; h[v] = ne; } char s[45][45]; int X[4] = {0,0,-1,1},Y[4] = {-1,1,0,0}; int lk[maxn],vis[maxn],N,M,sx,sy,n; int id[45][45],px[maxn],py[maxn],c[maxn],Out[maxn]; int win[maxn],ans[maxn],ansi; int bd[maxn][maxn]; void dfs(int x,int y,int co){id[x][y] = ++n; px[n] = x; py[n] = y; c[n] = co;int nx,ny;for (int k = 0; k < 4; k++){nx = x + X[k];ny = y + Y[k];if (nx < 1 || ny < 1 || nx > N || ny > M || s[nx][ny] == s[x][y])continue;if (id[nx][ny]){if (!bd[id[nx][ny]][id[x][y]]) build(id[x][y],id[nx][ny]);bd[id[nx][ny]][id[x][y]] = bd[id[x][y]][id[nx][ny]] = true;}else {dfs(nx,ny,co ^ 1);if (!bd[id[nx][ny]][id[x][y]]) build(id[x][y],id[nx][ny]);bd[id[nx][ny]][id[x][y]] = bd[id[x][y]][id[nx][ny]] = true;}} } bool find(int u){Redge(u) if (!vis[to = ed[k].to] && !Out[to]){vis[to] = true;if (!lk[to] || find(lk[to])){lk[to] = u; lk[u] = to;return true;}}return false; } int main(){N = read(); M = read();REP(i,N) scanf("%s",s[i] + 1);REP(i,N) REP(j,M) if (s[i][j] == '.'){sx = i,sy = j,s[i][j] = 'X';break;}dfs(sx,sy,0);int K = read() << 1,x,y,nx = sx,ny = sy,u;REP(i,n) if (!c[i]) cls(vis),find(i);REP(i,K){cls(vis);x = read(); y = read();Out[u = id[nx][ny]] = true;if (lk[u]){win[i] = !find(lk[u]);if (win[i]) lk[lk[u]] = 0;}else win[i] = 0;lk[u] = 0;nx = x; ny = y;}win[K] = true;for (int i = 1; i <= K; i += 2)if (win[i] && win[i + 1]) ans[++ansi] = (i + 1) >> 1;printf("%d\n",ansi);REP(i,ansi) printf("%d\n",ans[i]);return 0; }

轉載于:https://www.cnblogs.com/Mychael/p/9251088.html

總結

以上是生活随笔為你收集整理的BZOJ2437 [Noi2011]兔兔与蛋蛋 【博弈论 + 二分图匹配】的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。