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Posted on 2010-08-21 18:34 acronix 阅读(1644) 评论(0) 编辑 收藏 引用 所属分类: hzshuai收集的模板
匈牙利算法:
//poj_1469 /*==================================================*\ | 二分图匹配(匈牙利算法DFS 实现) | INIT: g[][]邻接矩阵; | 优点:实现简洁容易理解,适用于稠密图,DFS找增广路快。 | 找一条增广路的复杂度为O(E),最多找V条增广路,故时间复杂度为O(VE) \*==================================================*/ #include <cstdio> #include <memory.h>
bool g[110][310],flag,vis[310]; int match[310]; int p,n;
/************* 匈牙利算法**************/ int SearchPath(int u) { for (int v = 1; v <= n; v ++) if (g[u][v] && ! vis[v]) { vis[v] = true; if (match[v] == -1 || SearchPath(match[v])) { match[v] = u; return 1; } } return 0; } /**********************************/ int main() { int i,j,k,t,v,cnt; scanf("%d",&t); while (t--) { scanf("%d %d", &p, &n); for (i = 1; i <= p; i++) for (j = 1; j <= n; j++) g[i][j] = false; for (i = 1; i <= n; i++) match[i] = -1; flag = true; for (i = 1; i <= p; i++) { scanf("%d",&k); if (k == 0) flag = false; while (k--) { scanf("%d",&v); g[i][v] = true; } } if (flag) { cnt = 0; for (i = 1; i <= p; i++) { /***************/ memset(vis,false,sizeof(vis)); cnt += SearchPath(i); /*********************************/ } if (cnt == p) printf("YES\n"); else printf("NO\n"); } else printf("NO\n"); } return 0; }
Hopcroft-Carp 算法:
//poj_1469 /*==================================================*\ | 二分图匹配(Hopcroft-Carp 的算法) | INIT: g[][]邻接矩阵; | CALL: res = MaxMatch(); Nx, Ny要初始化!!! | Mx,My为match | 时间复杂度为O(V^0.5 E) \*==================================================*/ /***********************Hopcroft-Carp 算法****************************************/ #include <cstdio> #include <memory.h> #include <queue> using namespace std;
const int MAXN = 310; const int INF = 1 << 28; bool flag; int p,n; int Mx[MAXN], My[MAXN], Nx, Ny; int dx[MAXN], dy[MAXN], dis; bool vst[MAXN],g[110][310]; bool searchP(void) { queue <int> Q; dis = INF; memset(dx, -1, sizeof(dx)); memset(dy, -1, sizeof(dy)); for (int i = 1; i <= Nx; i++) if (Mx[i] == -1){ Q.push(i); dx[i] = 0; } while (!Q.empty()) { int u = Q.front(); Q.pop(); if (dx[u] > dis) break; for (int v = 1; v <= Ny; v++) if (g[u][v] && dy[v] == -1) { dy[v] = dx[u]+1; if (My[v] == -1) dis = dy[v]; else{ dx[My[v]] = dy[v]+1; Q.push(My[v]); } } } return dis != INF; }
bool DFS(int u){ for (int v = 1; v <= Ny; v++) if (!vst[v] && g[u][v] && dy[v] == dx[u]+1) { vst[v] = 1; if (My[v] != -1 && dy[v] == dis) continue; if (My[v] == -1 || DFS(My[v])) { My[v] = u; Mx[u] = v; return 1; } } return 0; }
int MaxMatch(void){ int res = 0; memset(Mx, -1, sizeof(Mx)); memset(My, -1, sizeof(My)); while (searchP()) { memset(vst, 0, sizeof(vst)); for (int i = 1; i <= Nx; i++) if (Mx[i] == -1 && DFS(i)) res++; } return res; }
/**********************************************************************/ int main() { int i,j,k,t,v,cnt; scanf("%d",&t); while (t--) { scanf("%d %d", &p, &n); for (i = 1; i <= p; i++) for (j = 1; j <= n; j++) g[i][j] = false; flag = true; for (i = 1; i <= p; i++) { scanf("%d",&k); if (k == 0) flag = false; while (k--) { scanf("%d",&v); g[i][v] = true; } } Nx = p; Ny = n; if (flag) { cnt = MaxMatch(); if (cnt == p) printf("YES\n"); else printf("NO\n"); } else printf("NO\n"); } return 0; }
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