n个花, m个花瓶, 每个花放到一个花瓶里会产生一个值w[i][j], 一个花只能放到一个花瓶里, 一个花瓶只能放一个花, 求产生的最大值。
带权二分图模板。
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; const int M = 105; bool sx[M], sy[M]; int match[M], w[M][M], n, m, d, lx[M], ly[M]; void init () { memset (w, 0, sizeof(w)); } bool dfs (int u) { int v; sx[u] = true; for (v = 0; v < m; v++) { if (!sy[v] && lx[u]+ly[v]==w[u][v]) { sy[v] = true; if (match[v] == -1 || dfs (match[v])) { match[v] = u; return true; } } } return false; } int KM () { int i, j, k, sum = 0; memset (ly, 0, sizeof(ly)); for (i = 0; i < n; i++) { lx[i] = -inf; for (j = 0; j < m; j++) if (lx[i] < w[i][j]) lx[i] = w[i][j]; } memset (match, -1, sizeof(match)); for (i = 0; i < n; i++) { while (1) { memset (sx, false, sizeof(sx)); memset (sy, false, sizeof(sy)); if (dfs (i)) break; d = inf; for (j = 0; j < n; j++) if (sx[j]) for (k = 0; k < m; k++) if (!sy[k]) d = min (d, lx[j]+ly[k]-w[j][k]); if (d == inf) return -1; for (j = 0; j < n; j++) if (sx[j]) lx[j] -= d; for (j = 0; j < m; j++) if (sy[j]) ly[j] += d; } } for (i = 0; i < m; i++) if (match[i] > -1) sum += w[match[i]][i]; return sum; } int main() { cin>>n>>m; for(int i = 0; i<n; i++) { for(int j = 0; j<m; j++) { scanf("%d", &w[i][j]); } } int ans = KM(); cout<<ans<<endl; return 0; }