HDU 1533 miles algorithm (best match with minimum weight)

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Time limit:10000/5000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3299 Accepted Submission (s): 1674

Problem Descriptionon 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.

Inputthere 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.

Outputfor each test case, output one line with the single integer, which are the minimum amount, in dollars, your need to PA Y.

Sample Input2 2.mh.5 5HH.. m...............mm. H7 8...H ..... H....... H....mmmhmmmm ... H....... H....... H.... 0 0

Sample Output21028

Sourcepacific Northwest 2004 Topic meaning: Give a n*m map, ' m ' means people, ' H ' means the house, people each can go up and down to the neighboring units, each walk once the total cost of +1, each house can only accommodate one person, How much is the total cost when everyone walks to the house. Idea: People put left, the house put on the right, the Benquan value between people x[i] go to the house y[j] cost.  Weights into negative numbers, km find the best match, the answer is negative. Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <iostream>5#include <vector>6#include <queue>7#include <cmath>8#include <Set>9 using namespacestd;Ten  One #defineN 105 A #defineINF 999999999 -  - intMaxintXintY) {returnX>y?x:y;} the intMinintXintY) {returnX<y?x:y;} - intAbsintXintY) {returnx<0?-x:x;} -  - structKM { +     intN, M;  -     intG[n][n];  +     intLx[n], ly[n], slack[n];  A     intLeft[n], right[n];  at     BOOLS[n], t[n];  -    -     voidInitintNintm) { -          This->n =N;  -          This->m =m;  -Memset (G,0,sizeof(g));  in     }   -    to     voidAdd_edge (intUintVintval) {   +G[U][V] + =Val;  -     }   the    *     BOOLDfsinti) { $S[i] =true; Panax Notoginseng          for(intj =0; J < M; J + +) {   -             if(T[j])Continue;  the             intTMP = Lx[i] + ly[j]-G[i][j];  +             if(!tmp) {   AT[J] =true;  the                 if(Left[j] = =-1||DFS (Left[j])) {   +LEFT[J] =i;  -Right[i] =J;  $                     return true;  $                 }   -}ElseSLACK[J] =min (slack[j], TMP);  -         }   the         return false;  -     }  Wuyi    the     voidUpdate () { -         intA =inf;  Wu          for(inti =0; I < m; i++)   -             if(! T[i]) A =min (A, slack[i]);  About          for(inti =0; I < n; i++)   $             if(S[i]) Lx[i]-=A;  -          for(inti =0; I < m; i++)   -             if(T[i]) Ly[i] + =A;  -     }   A    +     intkm () { theMemset (left,-1,sizeof(left));  -Memset (right,-1,sizeof(right));  $memset (Ly,0,sizeof(Ly));  the          for(inti =0; I < n; i++) {   theLx[i] =-inf;  the              for(intj =0; J < M; J + +)   theLx[i] =Max (Lx[i], g[i][j]);  -         }   in          for(inti =0; I < n; i++) {   the              for(intj =0; J < M; J + +) Slack[j] =inf;  the              while(1) {   Aboutmemset (S),false,sizeof(S));  thememset (T,false,sizeof(T));  the                 if(Dfs (i)) Break;  the                 Elseupdate ();  +             }   -         }   the         intAns =0; Bayi          for(inti =0; I < n; i++) {   theAns + =G[i][right[i]];  the         }   -         returnans;  -     }   the }kmm; the  the intN, M; the CharMap[n][n]; - intG[n][n]; the  the structnode{ the     intx, y;94 node () {} theNodeintAintb) { thex=A; they=b;98     } About }a[n], b[n]; - 101 Main ()102 {103     intI, J, K;104     intNX, NY; the      while(SCANF ("%d%d", &n,&m) = =2){106         if(n==0&&m==0) Break;107         108          for(i=0; i<n;i++) scanf ("%s", Map[i]);109nx=ny=0; theMemset (G,0,sizeof(g));111          for(i=0; i<n;i++){ the              for(j=0; j<m;j++){113                 if(map[i][j]=='m') a[nx++]=node (i,j); the                 if(map[i][j]=='H') b[ny++]=node (i,j); the             } the         }117 Kmm.init (nx,ny);118          for(i=0; i<nx;i++){119              for(j=0; j<ny;j++){ -Kmm.add_edge (i,j,-(ABS (a[i].x-b[j].x) +abs (a[i].y-( b[j].y) ));121             }122         }123printf"%d\n",-kmm.km ());124     } the}

HDU 1533 miles algorithm (best match with minimum weight)