HDU 5155 Harry and Magic Box (DP)

Harry and Magic Box

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): Accepted submission (s): 44

Problem Descriptionone Day, Harry got a magical box. The box is made of n*m grids. There is sparking jewel in some grids. But the top and bottom of the box are locked by amazing magic, so Harry can ' t see the inside from the top or bottom. However, four sides of the box is transparent, so Harry can see the inside from the four sides. Seeing from the left of the box, Harry finds each row is shining (it means each row have at least one jewel). And seeing from the front of the box, each column is shining (it means each column have at least one jewel). Harry wants to know how many kinds of jewel ' s distribution is there in the box. And the answer is too large, you should output the answer mod 1000000007.

Inputthere is several test cases.
For each test case,there is integers n and m indicating the size of the box. 0≤n,m≤

Outputfor Each test case, just output one line, contains an integer indicating the answer.

Sample INPUT1 12 22 3

Sample Output1725 HintThere is 7 possible arrangements for the second Test case.  They are:11 one-on-one-one-1011-one-one-one-01Assume-A-grids is ' 1 ' when it contains a jewel otherwise not. Set DP[I][J] indicates that the number of J columns has been selected in the previous I row.

#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<queue>#include<map>#include<Set>#include<stack>#include<algorithm>using namespacestd;#defineRoot 1,n,1#defineLson l,mid,rt<<1#defineRson mid+1,r,rt<<1|1#defineLR rt<<1#defineRR rt<<1|1typedefLong LongLl;typedef pair<int,int>PII;#defineX First#defineY SecondConst intOO = 1e9+7;Const DoublePI = ACOs (-1.0);Const DoubleEPS = 1e-6 ;Const intN = the ;Const intMoD = 1e9+7; LL Dp[n][n], f[n], c[n][n];intN, M;voidinit () {f[0] =1 ;  for(inti =1; I <= -; ++i) F[i] = f[i-1] *2 ;  for(inti =0; i < N; ++i) {c[i][0] = C[i][i] =1 ;  for(intj =1; J < I; ++j) {C[i][j]= (c[i-1][J] + c[i-1][j-1] ) %MoD; }    }}voidRun () {memset (DP,0,sizeofDP); dp[0][0] =1 ;  for(inti =0; I < n; ++i) {         for(intj =0; J <= M; ++j) {             for(intK =0; J + k <= m; k++) {LL cnt= F[j];if(!k) cnt--; CNT%=MoD; Dp[i+1][j+k] + = (dp[i][j] * CNT% mod * c[m-j][k])%MoD; Dp[i+1][J+K]%=MoD; }}} printf ("%i64d\n", Dp[n][m]);}intMain () {#ifdef LOCAL freopen ("In.txt","R", stdin); #endif //LOCALinit ();  while(~SCANF ("%d%d",&n,&m)) Run ();}

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HDU 5155 Harry and Magic Box (DP)