UVA 11082 Matrix decompressing "Max Flow"

Only see the topic words ~ ~ How can not see it is the network flow of the topic said ~ ~ ~ ~

Building the map is amazing ~ ~

At first I did not understand---later looked at this article--

Http://www.cnblogs.com/AOQNRMGYXLMV/p/4280727.html

Establish source Point st = 0, Meeting point ed = r+c

Because the range of positive integers is 1 to 20, and the flow can be 0, so first subtract each number inside the matrix by 1, and then add 1 to the final output answer.

Think of each row as a node x, numbered 1 to R

Consider each column as a node y, numbered r+1 through R+c

St to x Edge, capacity Ai '-C

Y to Ed Edge, Volume bi '-R

X-to-y edge with a capacity of 19

Just use it again, dinic.

The traffic from node XI to YJ is the value after lattice (i,j)-1

The beginning of the sentence was very, very incomprehensible-------

Then I searched the puzzle.

Because the YJ traffic comes from each X

Similarly, the flow of Xi flows to each Y, so (i,j) the flow source of this lattice is XI

Probably like this picture.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <queue>6 using namespacestd;7 8 Const intMAXN =10005;9 Const intINF = (1<< -) -1;Ten  One structedge{ A     intv,next,c; - }E[MAXN]; -  the intst,ed,lev[maxn],first[maxn],now[maxn],ecnt; - intr,c; -  - voidinit () { +memset (first,-1,sizeof(first)); -ECNT =0; + } A  at voidAddedges (intUintVintc) { -E[ecnt].next =First[u]; -E[ECNT].V =v; -E[ECNT].C =C; -First[u] = ecnt++; -      inE[ecnt].next =First[v]; -E[ECNT].V =u; toE[ECNT].C =0; +FIRST[V] = ecnt++; - } the  * BOOLBFs () { $queue<int>Q;Panax Notoginseng      while(!q.empty ()) Q.pop (); - Q.push (ST); thememset (lev,-1,sizeof(Lev)); +LEV[ST] =0; A      while(!Q.empty ()) { the         intx =Q.front (); Q.pop (); +          for(inti = First[x];~i;i =E[i].next) { -             intv =e[i].v; $             if(Lev[v] <0&& e[i].c >0){ $LEV[V] = Lev[x] +1; - Q.push (v); -             } the         } -     }Wuyi     returnLev[ed]! =-1; the } -  Wu intDfsintPintMinf) { -     if(p = = Ed | | minf = =0)returnMinf; About      for(int&i = Now[p];~i;i =E[i].next) { $         intv =e[i].v; -         if(Lev[v] = = Lev[p] +1&& e[i].c >0){ -             intD =Dfs (V,min (e[i].c,minf)); -             if(D >0){ AE[I].C-=D; +e[i^1].C + =D; the                 returnD; -             } $         } the     } the     return 0; the } the  - intDinic () { in     intMax_flow =0, p1; the      while(BFS ()) { thememcpy (Now,first,sizeof(first)); About          while((P1 = dfs (st,inf)) >0) theMax_flow + =P1; the     } the     returnMax_flow; + } -  the intans[ -][ -];Bayi  the intMain () { the     intT; -scanf"%d",&T); -     intKase =0; the      while(t--){ thescanf"%d%d",&r,&c); the init (); theSt =0; ed = r+c+1; -          the         intLast =0, cur;  the          for(inti =1; I <= r;i++){ thescanf"%d",&cur);94Addedges (st,i,cur-last-c); theLast =cur; the         } the         98Last =0; About          for(inti =1; I <= c;i++){ -scanf"%d",&cur);101Addedges (i + r,ed,cur-last-R);102Last =cur;103         }104          the          for(inti =1; I <= r;i++)106              for(intj =1; J <= c;j++) addedges (I,j+r, +);107         108         intres =dinic ();109          the          for(intU =1; u <= r;u++){111              for(inti = First[u];~i;i =E[i].next) { the                 intv =e[i].v;113ANS[U][V-R] = +-e[i].c; the             } the         } the         117printf"Matrix%d\n",++Kase);118          for(inti =1; I <= r;i++){119printf"%d", ans[i][1] +1); -              for(intj =2; J <= c;j++)121printf"%d", Ans[i][j] +1);122printf"\ n");123         }124         if(T) puts (""); the     }126     return 0;127}

View Code

UVA 11082 Matrix decompressing "Max Flow"