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NC51316. Going Home

描述

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. 
Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point. 

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

输入描述

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

输出描述

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

示例1

输入: 

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

输出: 

2
10
28

原站题解

上次编辑到这里，代码来自缓存 点击恢复默认模板

C++(clang++ 11.0.1) 解法, 执行用时: 3ms, 内存消耗: 420K, 提交时间: 2023-08-08 21:57:33 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define x first
#define y second
typedef pair<int,int>PII;
const int N=110;
const int INF=0x3f3f3f3f;
int n,m,P,H;
char s[N][N];
PII p[N],h[N];
int a[N][N];
int lx[N],ly[N];
int match[N];
bool vx[N],vy[N];
bool dfs(int u){
	vx[u]=1;
	for(int i=1;i<=H;i++){
		if(vy[i] || lx[u]+ly[i]!=a[u][i]) continue;
		vy[i]=1;
		if(match[i]==-1 || dfs(match[i])){
			match[i]=u;
			return true;
		}
	}
	return false;
}
int main(){
	while(scanf("%d%d",&n,&m),n&&m){
		P=H=0;
		for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++){
				if(s[i][j]=='m') p[++P]=make_pair(i,j);
				if(s[i][j]=='H') h[++H]=make_pair(i,j);
			}
		for(int i=1;i<=P;i++)
			for(int j=1;j<=H;j++)
				a[i][j]=-(abs(p[i].x-h[j].x)+abs(p[i].y-h[j].y));
		memset(lx,-0x3f,sizeof(lx));
		memset(ly,0,sizeof(ly));
		memset(match,-1,sizeof(match));
		for(int i=1;i<=P;i++){
			while(true){
				memset(vx,0,sizeof(vx));
				memset(vy,0,sizeof(vy));
				if(dfs(i)) break;
				int dt=INF;
				for(int j=1;j<=P;j++){
					if(!vx[j]) continue;
					for(int k=1;k<=H;k++){
						if(vy[k]) continue;
						dt=min(dt,lx[j]+ly[k]-a[j][k]);
					}
				}
				for(int j=1;j<=P;j++) if(vx[j]) lx[j]-=dt;
				for(int j=1;j<=H;j++) if(vy[j]) ly[j]+=dt;
			}
		}
		int ans=0;
		for(int i=1;i<=P;i++)
			ans=ans+lx[i]+ly[i];
		printf("%d\n",-ans);
	}
	return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 3ms, 内存消耗: 428K, 提交时间: 2022-10-16 15:24:28 

#include<iostream>
#include<cstring>
#define _for(i,L,R) for(int i=L;i<=R;++i)
//#define int long long

using namespace std;

using node=pair<int,int>;
#define x first
#define y second
const int N=1e2+5,INF=0x3f3f3f3f;
node h[N],p[N];
int visx[N], visy[N], lx[N], ly[N], link[N], g[N][N];
int n, m;
 
bool dfs(int x,int m)
{
	visx[x]=1;
	_for(i,1,m){
		if(!visy[i] and lx[x]+ly[i]==g[x][i]){
			visy[i]=1;
			if(link[i]==-1 or dfs(link[i],m)) return link[i]=x,1;
		}
	}
	return 0;
}
 
int KM(int n,int m)
{
	memset(ly,0,sizeof ly);
	memset(lx,-0x3f,sizeof lx);
	memset(link,-1,sizeof link);
	_for(i,1,n) _for(j,1,m) lx[i]=max(lx[i],g[i][j]);
	_for(i,1,n) while(true){
		memset(visx,0,sizeof visx);		
		memset(visy,0,sizeof visy);
		if(dfs(i,m)) break;
		int d=INF;
		_for(j,1,n) if(visx[j]) _for(k,1,m) if(!visy[k]) d=min(d,lx[j]+ly[k]-g[j][k]);
		if(d==INF) return -1;
		_for(j,1,n) if(visx[j]) lx[j]-=d;
		_for(j,1,n) if(visy[j]) ly[j]+=d; 
	}
	return 1;
}

void solve()
{
	while(~scanf("%d%d",&n,&m) and (n or m)){
		int nh=0,np=0;
		_for(i,1,n) _for(j,1,m){
			char c;
			cin>>c;
			if(c=='H') h[++nh]={i,j};
			if(c=='m') p[++np]={i,j}; 
		}
		_for(i,1,np) _for(j,1,nh) g[i][j]=-abs(h[j].x-p[i].x)-abs(h[j].y-p[i].y);
		KM(np,nh);
		int res=0;
		_for(i,1,nh) res+=-g[link[i]][i];
		printf("%d\n",res);
	}
}

signed main()
{
	int T = 1;
//	scanf("%d",&T);
	while(T--) solve();
	return 0;
}

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