POJ1469 COURSES binary map matching Hungarian algorithm

Original link http://www.cnblogs.com/zhouzhendong/p/8232649.html

Topic Portal-POJ1469Test Instructions Summary

In a large matrix, there are some obstacle points.

Now let's cover the non-barrier point with a small rectangle of 1*2, ask not to overwrite the obstacle point and not repeat the overlay, asking if all non-barrier points can be covered.

Solving

The topic is almost bare.

First notice that the two-tuple (x, y) of the read-in representation of the barrier points is a row and X is a column.

This toxic < poor assessment >

Then consider the algorithm. Readers can refer to the author's previous blog post.

For adjacent non-barrier points we built the edges back and forth. Then we give the original image according to a certain point of the Manhattan distance of the odd and even black and white dyeing, found that the black can only be connected to the white, white can only be connected to the black. So this is a two-part diagram.

Then we ran a gun to Hungary.

Because the side of the connection is back and forth, it is equivalent to repeating the calculation once, that is, the maximum number of matches doubled.

So it becomes the maximum number of non-barrier points to cover. The total number of direct and barrier points can be compared.

Note that using a two-dimensional array to store the graph will be tle (I will tle), the change to a group of analog linked list has passed. < poor evaluation ++>

Code

#include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm >using namespace Std;const int n=35,k=1100;struct gragh{int cnt,y[k*4],nxt[k*4],fst[k];void Clear () {Cnt=0;memset ( Fst,0,sizeof FST);} void Add (int a,int b) {y[++cnt]=b,nxt[cnt]=fst[a],fst[a]=cnt;}} G;int m,n,k,cnt,pl[n][n],tn[n][n],match[k],vis[k];bool Check (int x,int y) {return 1<=x&&x<=m&& 1<=y&&y<=n&&!pl[x][y];} BOOL Match (int x) {for (int i=g.fst[x];i;i=g.nxt[i]) {int y=g.y[i];if (!vis[y]) {vis[y]=1;if (!match[y]| | Match (Match[y])) {Match[y]=x;return 1;}}} return 0;} int Hungary () {int res=0;memset (match,0,sizeof match), for (int i=1;i<=cnt;i++) {memset (vis,0,sizeof Vis), if (Match (i )) res++;} return res;} int main () {while (~scanf ("%d%d%d", &m,&n,&k)) {memset (pl,0,sizeof pl); memset (tn,0,sizeof tn); for (int i=1, a,b;i<=k;i++) scanf ("%d%d", &b,&a), pl[a][b]=1;cnt=0;for (int i=1;i<=m;i++) for (int j=1;j<=n;j++) if ( !pl[i][J]) Tn[i][j]=++cnt;g.clear (); for (int i=1;i<=m;i++) for (int j=1;j<=n;j++) {if (Pl[i][j]) continue;if (check (i,j-1 ) G.add (Tn[i][j],tn[i][j-1]), if (check (i,j+1)) G.add (tn[i][j],tn[i][j+1]), if (check (i-1,j)) G.add (Tn[i][j],tn[i-1] [j]); if (check (i+1,j)) G.add (Tn[i][j],tn[i+1][j]);} Puts ((Hungary () ==cnt)? " YES ":" NO ");} return 0;}

　　

POJ1469 COURSES binary map matching Hungarian algorithm