The binary diagram of poj2446

Title:

id=2446 ">poj2446

Serie A champions: given a m*n matrix, in some places pits, and then 1*2 this article superimposed, repeatedly. Can put out the pit where all the other cover words output yes otherwise no

Analysis: There is a two-figure classic topic, of course, the difficulty is still a map, there is no idea, the early thought to be able to use (i-1) *m+j It every point in the matrix into a number, and then adjacent to the building map, Hungary, but do not know why is not correct? Ask the great God to explain, or understand not deep enough.

Very many people are based on their parity, because to be covered with 1*2 pieces of paper, then two values (I+J) must be an odd number of an even. And then to the figure of the odd number of points sequentially from 1 to start marking, adjacent to its label to build the map. Hungary, and relatively high speed. Positive Solution!

Because it must be an odd-numbered corresponding adjacent odd-numbered points. Then it is only possible to ask for the maximum matching of random odd or even numbers. The classic method of building a map.

Code:

#include <cstdio> #include <cstring> #include <string> #include <iostream> #include < Algorithm> #include <cmath>using namespace std;const int N = #define Del (x, y) memset (x,y,sizeof (×)) int map[        N][n],link[n],vis[n],vlink[n];int path[50][50];int n,m,t,tmp1,tmp2;bool dfs (int x) {for (int i=1; i<tmp2; i++) {            if (map[x][i]==1 && vis[i]==0) {vis[i]=1;                if (Link[i]==-1 | | dfs (LINK[I])) {link[i]=x;            return true; }}} return false;}    void Solve () {int ans=0;    Del (link,-1);    Del (vlink,-1);        for (int i=1; i<tmp1; i++) {Del (vis,0);    if (Dfs (i)) ans++;    }//printf ("%d\n", ans);    if (ans*2== (m*n-t)) printf ("yes\n"); else printf ("no\n");}    int main () {//freopen ("Input.txt", "R", stdin);        while (~SCANF ("%d%d", &n,&m)) {Del (path,0);        int x, y;     scanf ("%d", &t);           for (int i=0; i<t; i++) {scanf ("%d%d", &x,&y);        Path[y][x]=-1;        } tmp1=1,tmp2=1;                for (int i=1;i<=n;i++) {for (int j=1;j<=n;j++) {if (path[i][j]==0)                    {if ((i+j)%2==0) path[i][j]=tmp1++;                else path[i][j]=tmp2++;        }}} Del (map,0); for (int i=1, i<=n; i++) {for (int j=1; j<=m; J + +) {if (Path[i][j]!=-1 &A mp;& (i+j)%2==1) {if (path[i-1][j]>=1) map[path[i-1][j]][p                    Ath[i][j]]=1;                    if (path[i+1][j]>=1) map[path[i+1][j]][path[i][j]]=1;                    if (path[i][j-1]>=1) map[path[i][j-1]][path[i][j]]=1;               if (path[i][j+1]>=1)         Map[path[i][j+1]][path[i][j]]=1;    }}} solve (); } return 0;}

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The binary diagram of poj2446