HDU 5113 Black and white, color dyeing, tips

Black and WhiteTime limit:2000/2000 MS (java/others) Memory limit:512000/512000 K (java/others)
Total submission (s): 485 Accepted Submission (s): 131
Special Judge

Problem DescriptionIn Mathematics, the four-color theorem, or the four-color map theorem, states, given any separatio N of a plane into contiguous regions, producing a figure called a map, no more than four colors is required to color the Regions of the map so, no, adjacent regions have the same color.
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In this problem, you has to solve the 4-color problem. Hey, I ' m just joking.

You is asked to solve a similar problem:

The color an NXM chessboard with K colors numbered from 1 to K such, that no, and adjacent cells has the same color (both cell S is adjacent if they share an edge). The i-th color should is used in exactly CI cells.

Matt hopes you can tell him a possible coloring.
Inputthe first line contains only one integer T (1≤t≤5000), which indicates the number of test cases.

For each test case, the first line contains three integers:n, M, K (0 < N, m≤5, 0 < K≤NXM).

The second line contains K integers ci (ci > 0), denoting the number of cells where the i-th color should is used.

It ' s guaranteed that C1 + C2 + + CK = Nxm.

Outputfor Each test case, the first line contains ' case #x: ', where x is the case number (starting from 1).

In the second line, output "No" if there is no coloring satisfying the requirements. Otherwise, Output "YES" in the one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells.

If There is multiple solutions, output any of them.
Sample Input

41 5 24 13 3 41 2 2 42 3 32 2 23 2 32 2 2

 
Sample Output

Case #1: nocase #2: YES4 3 1 3 4Case #3: YES1 2 3 1Case #4: YES1 22 33 1

Technical Solution:

First black-and-white marks the board, and then divides all the color categories into three categories less than or equal to (n*m+1)/2. A1 > A2 > a3. The inability to divide is no solution.

Then the A1 class from the top down to fill the black mark of the lattice, A2 class from the bottom to fill the white mark of the lattice, the remaining lattice with A3 class (after not strictly prove A3 class will not be adjacent).

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < Cmath> #include <vector>using namespace std;const int maxn=51;int i,j,k;int n,m,nm,nm1,nm2,bj;int a[maxn],wz1[ Maxn],ans[maxn];int b[maxn],wz2[maxn],col[maxn];int line[4][maxn],gs[4];int cmp (int x,int y) {return gs[x]>gs[y];}    int main () {int t,q;    scanf ("%d", &t);        for (int. Ca=1;ca<=t;++ca) {printf ("Case #%d:\n", CA);        scanf ("%d%d%d", &n,&m,&q);        for (i=1;i<=q;++i) {scanf ("%d", &a[i]);        } nm=n*m;        nm1= (nm+1) >>1;                nm2=nm>>1;        int l1=0,l2=0;        col[0]=0;        for (int i=2;i<=m;++i) col[i]=1-col[i-1];        for (int i=m+1;i<=nm;++i) col[i]=1-col[i-m];            for (int i=1;i<=nm;++i) {if (col[i]) wz2[++l2]=i;        else wz1[++l1]=i;        } gs[1]=gs[2]=gs[3]=0;        for (i=1;i<4;++i) b[i]=i; for (i=1;i<=q && gs[1]+a[i]<=nm1;++i) {while (a[i]--) line[1][++gs[1]]=i;        } for (; i<=q && gs[2]+a[i]<=nm1;++i) {while (a[i]--) line[2][++gs[2]]=i;        } for (; i<=q && gs[3]+a[i]<=nm1;++i) {while (a[i]--) line[3][++gs[3]]=i;            } if (i<=q) {puts ("NO");        Continue        } puts ("YES");        Sort (b+1,b+4,cmp);        for (I=1;i<=nm1-gs[b[1]];++i) {ans[wz1[i]]=line[b[3]][i];        } k=i;        for (J=1;j<=gs[b[1]];++j,++i) {ans[wz1[i]]=line[b[1]][j];        } for (i=nm2;i>gs[b[2]];--i,++k) {ans[wz2[i]]=line[b[3]][k];        } for (; i>0;--i) {ans[wz2[i]]=line[b[2]][i];                } for (int i=0;i<n;++i) {for (int j=0;j<m;++j) {if (j) printf ("");            printf ("%d", ans[i*m+1+j]);        } puts (""); }}//SYstem ("pause"); return 0;}

HDU 5113 Black and white, color dyeing, tips