http://acm.hdu.edu.cn/showproblem.php?pid=1533
http://poj.org/problem?id=2195
C++版本一
题解：最小费用最大流
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200+10;
const int M=100+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int s,t,n,m,k,p,l,r,u,v;
int ans,cnth,cntm,flag,temp,sum,minz,maxflow;
int dis[N];
bool vis[N];
char str[N][N];
struct Node{
int x,y;
Node(){};
Node(int X,int Y):x(X),y(Y){}
}home[N],man[N];
struct node{
int u,v,c,w;
node(){};
node(int form,int to,int cap,int w):u(form),v(to),c(cap),w(w){}
};
vector<node>edge;
vector<int> G[N];
void Addedge(int u,int v,int cap,int w){
edge.push_back({u,v,cap,w});
edge.push_back({v,u,0,-w});
int sz=edge.size();
G[u].push_back(sz-2);
G[v].push_back(sz-1);
}
bool bfs(int u){
//memset(dis,-1,sizeof(dis));
for(int i=0;i<=t;i++)dis[i]=INF,vis[i]=0;
dis[u]=0;
queue<int>q;
q.push(u);
vis[u]=1;
while(!q.empty()){
int u=q.front();
q.pop();
vis[u]=0;
for(int i=0;i<G[u].size();i++){
node e=edge[G[u][i]];//cout<<u<<" "<<e.v<<endl;
if(dis[e.v]>dis[u]+e.w&&e.c>0){
dis[e.v]=dis[u]+e.w;
if(!vis[e.v]){
vis[e.v]=1;
q.push(e.v);
}
}
}
}
return dis[t]!=INF;
}
int dfs(int u,int flow){
vis[u]=1;
if(u==t)
return flow;
int now;
for(int i=0;i<G[u].size();i++){
node e=edge[G[u][i]];//cout<<ans<<endl;
if((!vis[e.v]||e.v==t)&&e.c>0&&dis[u]+e.w==dis[e.v]&&(now=dfs(e.v,min(flow,e.c)))){
ans+=e.w*now;
//cout<<now<<endl;
edge[G[u][i]].c-=now;
edge[G[u][i]^1].c+=now;
return now;
}
}
return 0;
}
void dinic(){
while(bfs(s)){
int res=0;
//memset(vis,0,sizeof(vis));
while((res=dfs(s,INF))){
maxflow+=res;
memset(vis,0,sizeof(vis));
}
}
}
void init(){
for(int i=0;i<=t;i++)G[i].clear();
edge.clear();
ans=0;
maxflow=0;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
while(~scanf("%d%d",&n,&m)&&n+m){
cnth=cntm=0;
for(int i=1;i<=n;i++){
scanf("%s",str[i]+1);
for(int j=1;j<=m;j++){
if(str[i][j]=='H')home[++cnth]=Node(i,j);
if(str[i][j]=='m')man[++cntm]=Node(i,j);
}
}
s=0;
t=cnth+cntm+1;
init();
for(int i = 1; i <= cnth; i++){//建图
for(int j = 1; j <= cntm; j++){
int dist= abs(home[i].x - man[j].x) + abs(home[i].y - man[j].y);
Addedge(i,j+cnth,1,dist);
//Addedge(j+cnth,i,0,-dist);
}
}
for(int i=1;i<=cnth;i++){
Addedge(s,i,1,0);
//Addedge(i,s,0,0);
}
for(int i=cnth+1;i<t;i++){
//Addedge(t,i,0,0);
Addedge(i,t,1,0);
}
//cout<<s<<" "<<t<<endl;
dinic();
cout<<ans<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
题解：带权二分图的最优匹配
每个man和house建立带权二分图，曼哈顿距离就是边的值，这里要求最小费用，也就是二分图最小权值匹配，但是KM算法求的是二分图最大权值匹配，所以此处用边的负数求最优匹配，求出来的答案的负数就是最小权匹配。
注意：题目说house最多100，但是没有说明man的范围，所以man应该最多100*100。
应该用house为二分图的X部，因为算法复杂度和X部点数有关，所以用点数少的house为X部
因为存的是负数，在预处理X部的顶标值初始化应该是-INF，不能是0
参考文章：KM算法
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100+10;
const int M=100+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnth,cntm,flag,temp,sum,minz;
int Map[N][N],cx[N],cy[N],wx[N],wy[N];
bool visx[N],visy[N];
char str[N][N];
struct node{
int x,y;
node(){};
node(int X,int Y):x(X),y(Y){}
}home[M],man[M];
void init(){
cnth=0;cntm=0;
ans=0;
}
bool dfs(int u){
visx[u]=1;
for(int v=1;v<=cntm;v++){
if(!visy[v]){
int t=wx[u]+wy[v]-Map[u][v];
if(t==0){
visy[v]=1;
if(cy[v]==-1||dfs(cy[v])){
cx[u]=v;
cy[v]=u;
return 1;
}
}else if(t>0)minz=min(minz,t);
}
}
return 0;
}
int KM(){
memset(cx,-1,sizeof(cx));
memset(cy,-1,sizeof(cy));
memset(wx,-INF,sizeof(wx));
memset(wy,0,sizeof(wy));
for(int i = 1; i <= cnth; i++){//建图
for(int j = 1; j <= cntm; j++){
wx[i]=max(wx[i],Map[i][j]);
}
}
for(int i=1;i<=cnth;i++){
while(1){
minz=INF;
memset(visx, 0, sizeof(visx));
memset(visy, 0, sizeof(visy));
if(dfs(i))break;
for(int j = 1; j <= cnth; j++)
if(visx[j])wx[j] -= minz;
for(int j = 1; j <= cntm; j++)
if(visy[j])wy[j] += minz;
}
}
int res=0;
for(int i=1;i<=cnth;i++)
if(cx[i]!=-1)res+=Map[i][cx[i]];
return res;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
while(~scanf("%d%d",&n,&m)&&n+m){
init();
for(int i=1;i<=n;i++){
scanf("%s",str[i]+1);
for(int j=1;j<=m;j++){
if(str[i][j]=='H')home[++cnth]=node(i,j);
if(str[i][j]=='m')man[++cntm]=node(i,j);
}
}
for(int i = 1; i <= cnth; i++){//建图
for(int j = 1; j <= cntm; j++){
Map[i][j] = -abs(home[i].x - man[j].x) - abs(home[i].y - man[j].y);
}
}
cout<<(-KM())<<endl;
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本三
题解：最小费用最大流
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define INF 0x3f3f3f3f
#define FI first
#define SE second
#define MP make_pair
#define PI pair<int,int>
#define lson l,m,rt<<1,ls,rs
#define rson m+1,r,rt<<1|1,ls,rs
#define test printf("here!!!\n")
using namespace std;
const int mx=100+10;//最大流的情况下使他最小费用
int n,m;
char s[mx];
int st_p,to_p;
struct EG
{
int to;
int c;
int w;
int nxt;
}edge[mx*mx*2];
int head[mx*2];
int cnt=1;
int dis[mx*2];
int vis[mx*2];
struct Nd
{
int x,y;
}hs[mx],ms[mx];
int tot,tot2;
int ans;
//int maxflow;
void add(int st,int to,int c,int w)
{
edge[++cnt].to=to;
edge[cnt].c=c;
edge[cnt].w=w;
edge[cnt].nxt=head[st];
head[st]=cnt;
}
int Dinic_SPFA()
{
for (int i=0;i<=tot+tot2+1;++i) dis[i]=INF,vis[i]=0;
queue<int>q;
q.push(st_p);
vis[st_p]=1;
dis[st_p]=0;
while (!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
for (int i=head[u];i;i=edge[i].nxt)
{
int v=edge[i].to;
if (dis[v]>dis[u]+edge[i].w&&edge[i].c)
{
dis[v]=dis[u]+edge[i].w;
if (!vis[v])
{
q.push(v);
vis[v]=1;
}
}
}
}
return dis[to_p]!=INF;
}//判断有无增广路，并构造最短边权的增广路,同BFS
int Dinic_dfs(int u,int flow)
{
vis[u]=1;
if (u==to_p)
{
//maxflow+=flow;
return flow;
}
int used=0;
for (int i=head[u];i;i=edge[i].nxt)
{
int v=edge[i].to;
if ((!vis[v]||v==to_p)&&dis[v]==dis[u]+edge[i].w&&edge[i].c)//防止0边权导致的环
{
int rflow=Dinic_dfs(v,min(flow-used,edge[i].c));
if (rflow)
{
used+=rflow;
ans+=edge[i].w*rflow;
edge[i].c-=rflow;
edge[i^1].c+=rflow;
if (flow==used) break;
}
}
}
return used;
}
int Dinic()
{
while (Dinic_SPFA())
{
vis[to_p]=1;
while (vis[to_p])
{
memset(vis,0,sizeof(vis));
Dinic_dfs(st_p,INT_MAX);
}
}
}
int main()
{
while (~scanf("%d%d",&n,&m)&&n+m)
{
memset(head,0,sizeof(head));
cnt=1;
tot=tot2=0;
ans=0;
//maxflow=0;
for (int i=1;i<=n;++i)
{
scanf("%s",s);
for (int j=0;j<m;++j)
{
if (s[j]=='m') ms[++tot].x=i,ms[tot].y=j;
else if (s[j]=='H') hs[++tot2].x=i,hs[tot2].y=j;
}
}
for (int i=1;i<=tot;++i)
{
for (int j=1;j<=tot2;++j)
{
int disct=abs(ms[i].x-hs[j].x)+abs(ms[i].y-hs[j].y);
add(i,tot+j,1,disct);
add(tot+j,i,0,-disct);
}
}
st_p=0;
to_p=tot+tot2+1;
for (int i=1;i<=tot;++i)
{
add(st_p,i,1,0);
add(i,st_p,0,0);
}
for (int i=tot+1;i<to_p;++i)
{
add(i,to_p,1,0);
add(to_p,i,0,0);
}
Dinic();
printf("%d\n",ans);
}
}
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