狀壓DP。。。

Kings on a Chessboard

Time Limit:?10000msMemory Limit:?65535KBThis problem will be judged on UESTC. Original ID:?1851
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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\n",ans%MOD); ????} ????return?0; }

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

轉(zhuǎn)載于:https://www.cnblogs.com/CKboss/p/3350829.html

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