Black And White HDU5113 / UVAlive7059

Black And White HDU5113 / UVAlive7059

In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color.
— Wikipedia, the free encyclopedia

In this problem, you have to solve the 4-color problem. Hey, I’m just joking.
You are asked to solve a similar problem:
Color an N × M chessboard with K colors numbered from 1 to K such that no two adjacent cells have the same color (two cells are adjacent if they share an edge). The i-th color should be used in exactly c i cells.
Matt hopes you can tell him a possible coloring. 

Input

The 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 ≤ N × M ).
The second line contains K integers c i (c i > 0), denoting the number of cells where the i-th color should be used.
It’s guaranteed that c 1 + c 2 + · · · + c K = N × M .

Output

For 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 one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells.
If there are multiple solutions, output any of them.

Sample Input

4
1 5 2
4 1
3 3 4
1 2 2 4
2 3 3
2 2 2
3 2 3
2 2 2

Sample Output

Case #1:
NO
Case #2:
YES
4 3 4
2 1 2
4 3 4
Case #3:
YES
1 2 3
2 3 1
Case #4:
YES
1 2
2 3
3 1

Solution

构造法。
首先，假如max{c[i]}>(n*m+1)/2,那么无解，否则一定有解。
然后，可以先把n*m的矩形网格划分成两部分，并把c[i]从大到小排序。
显然，假如n>=3，而且c[i]里面的第一大的和第二大的依次填入可能冲突。易证，此时将第二大与最小交换一下填充次序就可以了
最后，先填矩形网格的第一个部分直到填满，再填第二部分。

Code

#include <bits/stdc++.h>
using namespace std;
int a[10][10],T,n,m,k,p;
struct data {
    int num,id;
} c[30];
bool cmp(data A,data B) {
    return A.num>B.num;
}
bool ok;
void solve() {
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=m; ++j)
            if ((i+j)%2==0&&a[i][j]==-1) {
                a[i][j]=c[p].id;
                c[p].num--;
                if (c[p].num==0) p++;
            }
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=m; ++j)
            if (a[i][j]==-1) {
                a[i][j]=c[p].id;
                c[p].num--;
                if (c[p].num==0) p++;
            }
}
void output() {
    printf("YES\n");
    if (ok) {
        for (int i=m; i>=1; --i) {
            for (int j=1; j<n; ++j)
                printf("%d ",a[j][i]);
            printf("%d\n",a[n][i]);
        }

    } else {
        for (int i=1; i<=n; ++i) {
            for (int j=1; j<m; ++j) printf("%d ",a[i][j]);
            printf("%d\n",a[i][m]);
        }
    }
    ok=0;
}
int main() {
    scanf("%d",&T);
    for (int I=1; I<=T; ++I) {
        printf("Case #%d:\n",I);
        scanf("%d%d%d",&n,&m,&k);
        memset(a,0,sizeof(a));
        for (int i=0; i<k; ++i) {
            scanf("%d",&c[i].num);
            c[i].id=i+1;
        }
        sort(c,c+k,cmp);
        if (c[0].num>(n*m+1)/2) {
            printf("NO\n");
        } else {
            if (k>=3) swap(c[1],c[k-1]);
            p=0;
            memset(a,-1,sizeof(a));
            if (n%2==0&&m%2!=0) {
                swap(n,m);
                ok=1;
            }
            solve();
            output();
        }
    }
    return 0;
}