Going Home
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21730 | Accepted: 10990 |
Description
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.
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.
Input
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.
Output
For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input
2 2 .m H. 5 5 HH..m ..... ..... ..... mm..H 7 8 ...H.... ...H.... ...H.... mmmHmmmm ...H.... ...H.... ...H.... 0 0
Sample Output
2 10 28
Source
题目意思：
N行M列，每人可分得一个房子，每人每走一步花费为1，求每人各回各家的最小花费。
解题思路：
这题有两种思路可破：KM算法和最小费用最大流算法。
（一）
套Kuhn-Munkras算法模板，建二分图求解。
下面详细讲讲建图的过程：
①扫描读入的图，分别标记出人M和房屋H的位置坐标；
②每个人与每间房屋一一对应，比如测试用例
5 5
HH..m
.....
.....
.....
mm..H那么分别求出AX AY AZ BX BY BZ CX CY CZ路径上对应花费的权值就是二分图中对应边的边权。
因为是求最小权，所以建图时边的权值取相反数，最后所得的ans取相反数。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 300
#define INF 0x3f3f3f3f
int nx,ny;//两边的点数
int g[maxn][maxn];//二分图描述
int linker[maxn],lx[maxn],ly[maxn];//y中各点匹配状态，x,y中的点标号
int slack[maxn];
bool visx[maxn],visy[maxn];
int ax[maxn],ay[maxn],bx[maxn],by[maxn];
char Map[maxn][maxn];
bool DFS(int x)
{
visx[x] = true;
for(int y = 1; y <= ny; y++)
{
if(visy[y])continue;
int tmp = lx[x] + ly[y] - g[x][y];
if(tmp == 0)
{
visy[y] = true;
if(linker[y] == -1 || DFS(linker[y]))
{
linker[y] = x;
return true;
}
}
else if(slack[y] > tmp)
slack[y] = tmp;
}
return false;
}
int KM()
{
memset(linker,-1,sizeof(linker));
memset(lx,0,sizeof(lx));
memset(ly,0,sizeof(ly));
for(int i = 0; i < nx; i++)
{
lx[i] = -INF;
for(int j = 1; j <= ny; j++)
if(g[i][j] > lx[i])
lx[i] = g[i][j];
}
for(int x = 1; x <= nx; x++)
{
for(int i = 1; i <= ny; i++)
slack[i] = INF;
while(true)
{
memset(visx,false,sizeof(visx));
memset(visy,false,sizeof(visy));
if(DFS(x))break;
int d = INF;
for(int i = 1; i <= ny; i++)
if(!visy[i] && d > slack[i])
d = slack[i];
for(int i = 1; i <= nx; i++)
if(visx[i])
lx[i] -= d;
for(int i = 1; i <= ny; i++)
{
if(visy[i])ly[i] += d;
else slack[i] -= d;
}
}
}
int res = 0;
for(int i = 1; i <= nx; i++)
if(linker[i] != -1)
res += g[linker[i]][i];
return -res;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
{
/*for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
scanf("%d",&g[i][j]);
nx = ny = n;*/
memset(g,0,sizeof g);
nx=ny=0;//二分图顶点的个数
for(int i=1; i<=n; i++)
scanf("%s",Map[i]+1);//读入原图
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
{
if(Map[i][j]=='m')
{
nx++;
ax[nx]=i;
ay[nx]=j;
}
else if(Map[i][j]=='H')
{
ny++;
bx[ny]=i;
by[ny]=j;
}
}
for(int i=1; i<=nx; i++)
for(int j=1; j<=ny; j++)
g[i][j]=-(abs(ax[i]-bx[j])+abs(ay[i]-by[j]));//建图
printf("%d\n",KM());
}
return 0;
}
/**
2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0
**/
（二）
套最小费用最大流算法模板，加超级源点和超级汇点网络流求解。
注意模板使用的时候，minCostMaxflow()的返回值。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 300
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)//加边
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)//找增广路
{
queue<int>q;
for(int i = 0; i <= t+2; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//int minCostMaxflow(int s,int t,int &cost)
int minCostMaxflow(int s,int t)//最小费用最大流
{
int flow = 0;
int ans=0;
//cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
//cost += edge[i].cost * Min;
}
ans+=dis[t];
flow += Min;
}
return ans;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
{
int nx,ny,ax[maxn],ay[maxn],bx[maxn],by[maxn];
char Map[maxn][maxn];
nx=ny=0;//二分图顶点的个数
for(int i=1; i<=n; i++)
scanf("%s",Map[i]+1);//读入原图
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
{
if(Map[i][j]=='m')
{
nx++;
ax[nx]=i;
ay[nx]=j;
}
else if(Map[i][j]=='H')
{
ny++;
bx[ny]=i;
by[ny]=j;
}
}
int s=0,t=nx+ny+1;//源点和汇点
init(t+1);//初始化
for(int i = 1; i <= nx; i++)
addedge(s,i,1,0);//源点向各个点加边
for(int i = nx+1; i <= nx+ny; i++)
addedge(i,t,1,0);//汇点向各个点加边
for(int i=1; i<=nx; i++)
for(int j=1; j<=ny; j++)
{
int va=(abs(ax[i]-bx[j])+abs(ay[i]-by[j]));
//cout<<i<<"->"<<nx+j<<"="<<va<<endl;
addedge(i,nx+j,1,va);//建图
}
printf("%d\n",minCostMaxflow(s,t));
}
return 0;
}
/**
2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0
**/
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