链接：

https://vjudge.net/problem/LightOJ-1246

题意：

You are given a rectangular board. You are asked to draw M horizontal lines and N vertical lines in that board, so that the whole board will be divided into (M+1) x (N+1) cells. So, there will be M+1 rows each of which will exactly contain N+1 cells or columns. The yth cell of xth row can be called as cell(x, y). The distance between two cells is the summation of row difference and column difference of those two cells. So, the distance between cell(x1, y1) and cell(x2, y2) is

|x1 - x2| + |y1 - y2|

For example, the distance between cell (2, 3) and cell (3, 2) is |2 - 3| + |3 - 2| = 1 + 1 = 2.

After that you have to color every cell of the board. For that you are given K different colors. To make the board more beautiful you have to make sure that no two cells having the same color can have odd distance between them. For example, if you color cell (3, 5) with red, you cannot color cell (5, 8) with red, as the distance between them is 5, which is odd. Note that you can keep some color unused, but you can‘t keep some cell uncolored.

You have to determine how many ways to color the board using those K colors.

思路:

将图分为两个部分，任意一个部分的点到另一个部分的任意一个点的距离都是奇数。
这样就转变成了一个DP,Dp[i][j]表示n个位置，放了j种颜色。
Dp[i][j] = Dp[i-1][j-1]j+Dp[i-1][j]j
再用组合数分配一下两个部分的颜色。

代码：

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

int n, m, k;
LL Dp[410][55];
LL c[55][55];

void Init()
{
    memset(c, 0, sizeof(c));
    for (int i = 0;i < 55;i++)
    {
        c[i][0] = 1;
        for (int j = 1;j <= i;j++)
        {
            c[i][j] = (c[i-1][j-1] + c[i-1][j])%MOD;
        }
    }
    for (int i = 1;i < 410;i++)
    {
        Dp[i][1] = 1;
        for (int j = 2;j < 55;j++)
        {
            Dp[i][j] = (Dp[i-1][j-1]*j+Dp[i-1][j]*j)%MOD;
        }
    }
}

int main()
{
    // freopen("test.in", "r", stdin);
    Init();
    int t, cas = 0;
    scanf("%d", &t);
    while(t--)
    {
        printf("Case %d:", ++cas);
        scanf("%d%d%d", &n, &m, &k);
        n++, m++;
        LL ans = 0;
        if (n == m && m == 1)
            ans = k;
        else
        {
            int sum1 = 0, sum2 = 0;
            sum1 = ((n+1)/2)*((m+1)/2)+(n/2)*(m/2);
            sum2 = n*m-sum1;
            for (int i = 1;i < k;i++)
            {
                for (int j = 1;i+j <= k;j++)
                    ans = (ans + c[k][i]*c[k-i][j]%MOD*Dp[sum1][i]%MOD*Dp[sum2][j]%MOD)%MOD;
            }
        }
        printf(" %lld\n", ans);
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/12019331.html