UESTC 1851 Kings on a Chessboard

状压DP。。。

Kings on a Chessboard

Time Limit: 10000ms

Memory Limit: 65535KB

This problem will be judged on UESTC. Original ID: 1851
64-bit integer IO format: %lld      Java class name: Main

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You are given a chessboard of size x * y and k identical kings, and are asked to place all the kings on the board such that no two kings can attack each other. Two kings can attack each other if they are horizontally, vertically or diagonally adjacent.
Write a computer program that calculates the number of possible arrangements of the k kings on the given chessboard. Since the number of feasible arrangements may be large, reduce the number modulo 1,000,000,007.

Input

The first line of the input consists of a single integer T, the number of test cases. Each of the following T lines consists of three integers x; y and k,separated by one space.

0 < T <= 50
2 <= x; y <= 15
1 <= k <= x*y

Output

For each test case, output the number of possibilities modulo 1,000,000,007.

Sample Input

4
8 8 1
7 7 16
7 7 7
3 7 15

Sample Output

64
1
2484382
0

Source

IDI Open 2013 Programming Contest

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int MOD=1000000007;

inline bool legal(int x,int y) {return x&y;}
long long int dp[16][1600][250];
int r,c,nums,state[1600],people[1600],kth;

bool isOK(int xia,int shang)
{
    int x=state[xia],y=state[shang];
    if(legal(x,y)) return false;
    if(legal(x<<1,y)||legal(x>>1,y)) return false;
    return true;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d%d%d",&r,&c,&kth);
        if(kth>(r+1)/2*(c+1)/2)
        {
            puts("0");
            continue;
        }
        if(c>r) swap(r,c);
        ///zuangtai
        nums=0;
        memset(state,0,sizeof(state));
        memset(people,0,sizeof(people));
        for(int i=0;i<(1<<c);i++)
        {
            if(legal(i,i<<1)||legal(i,i>>1)) continue;
            state[nums]=i;
            int k=i;
            while(k)
            {
                if(k&1) people[nums]++;
                k=k>>1;
            }
            nums++;
        }
        ///the firstline
        for(int i=0;i<nums;i++)
        {
            dp[1][people]=1;
        }
        for(int i=2;i<=r;i++)
        {
            for(int j=0;j<nums;j++)
            {
                for(int k=0;k<nums;k++)
                {
                    if(!isOK(j,k)) continue;
                    for(int l=people[k];l<250;l++)
                    {
                        if(l+people[j]<250)
                            dp[j][l+people[j]]=(dp[j][l+people[j]]+dp[i-1][k][l])%MOD;
                    }
                }
            }
        }
        long long int ans=0;
        for(int j=0;j<nums;j++)
        {
            ans=(ans+dp[j][kth])%MOD;
        }
        printf("%lld ",ans%MOD);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )