POJ 2195 Going Home (最佳完美匹配, 最小费用最大流)

题目类型  最佳完美匹配, 最小费用最大流

题目意思

给出一个最多 100 * 100 的字符矩阵 其中有若干个m和相同数量的H, 现在要使每个m都与一个不同的H配对,问最少的花费是多少

一次配对的花费是配对的两个字符的哈密顿距离

解题方法

用km算法求最佳完美匹配(即花费最小的完美匹配) 每个m点和所有的H点连一条权值为原花费*(-1)的边 然后求一次权值和最大的完美匹配即可

用最小费用最大流的方法做就是新建一个源点s和一个汇点t, s到所有的m连一条容量为1, 费用为0的边,所有的H到t连一条容量为1费用为0的边, 每个ｍ与H之间连一条容量为1费用为相应花费的边, 然后求一次最小费用最大流即可

参考代码 - 有疑问的地方在下方留言 看到会尽快回复的

km算法:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

const int maxn = 100 + 10;
const int INF = 1<<29;

char str[maxn][maxn];
vector<int>M, H;
int Lx[maxn], Ly[maxn], Left[maxn];
int s[maxn], t[maxn], w[maxn][maxn];
int sum;

bool match(int i, int n) {
	s[i] = true;
	for( int j=1; j<=n; j++ ) if(Lx[i]+Ly[j] == w[i][j] && !t[j]) {
		t[j] = true;
		sum += w[i][j];
		if(!Left[j] || match(Left[j], n)) {
			sum -= w[Left[j]][j];
			Left[j] = i;
			return true;
		}
		sum -= w[i][j];
	}
	return false;
}

void update(int n) {
	int a = 1<<30;
	for( int i=1; i<=n; i++ ) if(s[i]) 
		for( int j=1; j<=n; j++ ) if(!t[j]) 
			a = min(a, Lx[i]+Ly[j]-w[i][j]);
	for( int i=1; i<=n; i++ ) {
		if(s[i]) Lx[i] -= a;
		if(t[i]) Ly[i] += a;
	}
}

void km(int n) {
	for( int i=1; i<=n; i++ ) {
		Left[i] = Lx[i] = Ly[i] = 0;
		for( int j=1; j<=n; j++ ) {
			Lx[i] = max(Lx[i], w[i][j]);
		}
	}
	for( int i=1; i<=n; i++ ) {
		for( ; ;) {
			for( int j=1; j<=n; j++ ) s[j] = t[j] = 0;
			if(match(i, n)) break; else update(n);
		}
	}
}

int Abs(int x) { return x < 0 ? -x : x; }

int main() {
	//freopen("in", "r", stdin);
	int n, m, N;
	while(scanf("%d%d", &n, &m), n || m) {
		M.clear(), H.clear();
		for( int i=0; i<n; i++ ) {
			scanf("%s", str[i]);
			for( int j=0; j<m; j++ ) {
				if(str[i][j] == 'm') M.push_back(i*m+j);
				else if(str[i][j] == 'H') H.push_back(i*m+j);
			}
		}
		N = M.size();
		for( int i=0; i<N; i++ ) {
			int xi = M[i]/m, yi = M[i]%m;
			int ti = i + 1;
			Lx[ti] = -INF;
			for( int j=0; j<N; j++ ) {
				int xj = H[j]/m, yj = H[j]%m;
				int cost = Abs(xi-xj) + Abs(yi - yj);
				int tj = j + 1;
				w[ti][tj] = -cost;
				Lx[ti] = max(Lx[ti], -cost);
			}
		}
		sum = 0;
		km(N);
		printf("%d\n", -sum);
	}
	return 0;
}

最小费用最大流:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const int maxn = 100 + 10;
const int INF = 1<<29;

struct Edge {
	int from, to, cap, flow, cost;
	Edge(int _from, int _to, int _cap, int _flow, int _cost) : from(_from), to(_to), cap(_cap), flow(_flow), cost(_cost) {}
	Edge() {}
};

struct MCMF {
	int n, m, s, t;
	vector<Edge>edges;
	vector<int>G[maxn*maxn];
	int inq[maxn*2], d[maxn*2], p[maxn*2], a[maxn*2];
	void init(int n) {
		this->n = n;
		for( int i=0; i<n; i++ ) G[i].clear();
		edges.clear();
	}
	void AddEdge(int from, int to, int cap, int cost) {
		edges.push_back(Edge(from, to, cap, 0, cost));
		edges.push_back(Edge(to, from, 0, 0, -cost));
		m = edges.size();
		G[from].push_back(m-2);
		G[to].push_back(m-1);
	}

	bool BellmanFord(int s, int t, int & flow, int & cost) {
		for( int i=0; i<n; i++ ) d[i] = INF;
		memset(inq, 0, sizeof(inq));
		d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
		queue<int>Q;
		Q.push(s);
		while(!Q.empty()) {
			int u = Q.front(); Q.pop();
			inq[u] = 0;
			for( int i=0; i<G[u].size(); i++ ) {
				Edge & e = edges[G[u][i]];
				if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
					d[e.to] = d[u] + e.cost;
					p[e.to] = G[u][i];
					a[e.to] = min(a[u], e.cap - e.flow);
					if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
				}
			}
		}
		if(d[t] == INF) return false;
		flow += a[t];
		cost += d[t] * a[t];
		int u = t;
		while(u != s) {
			edges[p[u]].flow += a[t];
			edges[p[u]^1].flow -= a[t];
			u = edges[p[u]].from;
		}
		return true;
	}

	int Mincost(int s, int t) {
		int flow = 0, cost = 0;
		while(BellmanFord(s, t, flow, cost));
		return cost;
	}
}MC;

char str[maxn][maxn];

int abs(int x) { return x < 0 ? -x : x; }
int main() {
	freopen("in", "r", stdin);
	int n, m;
	while(scanf("%d%d", &n, &m), n || m) {
		vector<int>M, H;
		for( int i=0; i<n; i++ ) {
			scanf("%s", str[i]);
			for( int j=0; j<m; j++ ) {
				if(str[i][j] == 'm') M.push_back(i*m+j);
				else if(str[i][j] == 'H') H.push_back(i*m+j);
			}
		}
		int N = M.size();
		MC.init(2*N+2);
		for( int i=0; i<N; i++ ) {
			int xi = M[i]/m, yi = M[i]%m;
			for( int j=0; j<N; j++ ) {
				int xj = H[j]/m, yj = H[j]%m;
				int cost = abs(xi-xj) + abs(yi - yj);
				MC.AddEdge(i+1, N+1+j, 1, cost);
			}
		}
		for( int i=1; i<=N; i++ ) MC.AddEdge(0, i, 1, 0);
		for( int i=N+1; i<=2*N; i++ ) MC.AddEdge(i, 2*N+1, 1, 0);
		int ans = MC.Mincost(0, 2*N+1);
		printf("%d\n", ans);
	}
	return 0;
}