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;}