POJ2195 going home[Cost flow | Binary graph maximum weight matching]

going Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22088		Accepted: 11155

Description

On a grid map there is n little men and N houses. In each unit time, every little mans can move one unit step, either horizontally, or vertically, to a adjacent point. For each little mans, you need to pay a $ fee for every step he moves, until he enters a house. The task is complicated with the restriction, 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 Diffe Rent 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 ' indica TES There is a little man on this point.

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

Input

There is one or more test cases in the input. Each case starts with a line giving-integers n and m, where N is the number of rows of the map, and M are the number of Columns. The rest of the input would be N lines describing the map. Assume both N and M are between 2 and inclusive. There would be the same number of ' H ' s and ' M's on the map; And there'll is at the most houses. Input would terminate with 0 0 for N and M.

Output

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

Sample Input

2 2.mh.5 5HH.. m...............mm. H7 8...H ..... H....... H....mmmhmmmm ... H....... H....... H.... 0 0

Sample Output

21028

Source

Pacific Northwest 2004 each and every house has a double-edged, two-figure maximum-weight matchsolution with SPFA cost flow (or km)

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespaceStd;typedefLong Longll;Const intn=5005, m=1e6+5, inf=1e9;intRead () {CharC=getchar ();intx=0, f=1;  while(c<'0'|| C>'9'){if(c=='-') f=-1; C=GetChar ();}  while(c>='0'&&c<='9') {x=x*Ten+c-'0'; C=GetChar ();} returnx*F;}intn,m,s,t,n1,n2;Charss[ the];structdata{intx, y;} A[n],b[n];inlineintDis (data &a,data &b) {returnABS (a.x-b.x) +abs (a.y-b.y);}structedge{intV,ne,c,f,w;} E[m<<1];intCnt,h[n];inlinevoidInsintUintVintCintW) {CNT++; E[CNT].V=v;e[cnt].c=c;e[cnt].f=0; e[cnt].w=W; E[cnt].ne=h[u];h[u]=CNT; CNT++; E[CNT].V=u;e[cnt].c=0; e[cnt].f=0; e[cnt].w=-W; E[cnt].ne=h[v];h[v]=CNT;}voidbuild () {CNT=0; memset (H,0,sizeof(h)); S=0; t=n1+n2+1;  for(intI=1; i<=n1;i++)         for(intj=1; j<=n2;j++) ins (i,n1+j,1, Dis (a[i],b[j]));  for(intI=1; i<=n1;i++) ins (S,i,1,0);  for(intI=1; i<=n2;i++) ins (n1+i,t,1,0);}intD[n],q[n],head,tail,inq[n],pre[n],pos[n];inlinevoidLopint&AMP;X) {if(x==n) x=1;}BOOLSPFA () {memset (d,127,sizeof(d)); memset (INQ,0,sizeof(INQ)); Head=tail=1; D[s]=0; inq[s]=1; q[tail++]=s; Pre[t]=-1;  while(head!=tail) {        intu=q[head++];inq[u]=0; Lop (head);  for(intI=h[u];i;i=e[i].ne) {            intv=e[i].v,w=E[I].W; if(d[v]>d[u]+w&&e[i].c>e[i].f) {D[v]=d[u]+W; PRE[V]=u;pos[v]=i; if(!inq[v]) q[tail++]=v,inq[v]=1, lop (tail); }        }    }    returnpre[t]!=-1;}intMCMF () {intflow=0, cost=0;  while(SPFA ()) {intf=INF;  for(intI=t;i!=s;i=pre[i]) f=min (f,e[pos[i]].c-e[pos[i]].f); Flow+=f;cost+=d[t]*F;  for(intI=t;i!=s;i=Pre[i]) {E[POS[I]].F+=F; e[((pos[i)-1)^1)+1].f-=F; }    }    returnCost ;}intMainintargcConst Char*argv[]) {     while(true) {N1=n2=0; N=read (); m=read (); if(n==0&&m==0) Break;  for(intI=1; i<=n;i++) {scanf ("%s", ss+1);  for(intj=1; j<=m;j++){                if(ss[j]=='H') a[++n1]=(data) {I,J}; if(ss[j]=='m') b[++n2]=(data) {I,J};        }} build (); printf ("%d\n", MCMF ()); }}

POJ2195 going home[cost flow | Binary graph maximum weight matching]