全局最小割模版 n^3

Posted 2020-09-18 jhcelue

//点标从0-n-1, 開始时先init 复杂度n^3
//对于边(u,v,flow):
//g[u][v]+=flow;
//g[v][u]+=flow;
typedef long long ll;
const int N = 305;
const ll inf = 1e18;
ll g[N][N], w[N];
int a[N], v[N], na[N];
ll mincut(int n) {
	int i, j, pv, zj;
	ll best = inf;
	for(i = 0; i < n; i ++) v[i] = i;

	while(n > 1) {
		for(a[v[0]] = 1, i = 1; i < n; i ++) {
			a[v[i]] = 0;
			na[i-1] = i;
			w[i] = g[v[0]][v[i]];
		}
		for(pv = v[0], i = 1; i < n; i ++) {
			for(zj = -1, j = 1; j < n; j ++)
				if(!a[v[j]] && (zj < 0 || w[j] > w[zj])) zj = j;

			a[v[zj]] = 1;
			if(i == n-1) {
				if(best > w[zj]) best = w[zj];
				for(i = 0; i < n; i ++) {
					g[v[i]][pv] = g[pv][v[i]] += g[v[zj]][v[i]];
				}
				v[zj] = v[--n];
				break;
			}
			pv = v[zj];
			for(j = 1; j < n; j ++) if(!a[v[j]])
				w[j] += g[v[zj]][v[j]];
		}
	}
	return best;
}
void init(int n){
    for(int i = 0; i < n; i ++)
        for(int j = 0; j < n; j ++)
            g[i][j] = g[j][i] = 0;
}