(CodeForces
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(CodeForces - 598E)Chocolate Bar
time limit per test:2 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output
You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares, so you may need to break the chocolate bar.
In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.
For example, if you have a chocolate bar consisting of 2 × 3 unit squares then you can break it horizontally and get two 1 × 3 pieces (the cost of such breaking is 32 = 9), or you can break it vertically in two ways and get two pieces: 2 × 1 and 2 × 2 (the cost of such breaking is 22 = 4).
For several given values n, m and k find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to k squares. The remaining n·m - k squares are not necessarily form a single rectangular piece.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 40910) — the number of values n, m and k to process.
Each of the next t lines contains three integers n, m and k (1 ≤ n, m ≤ 30, 1 ≤ k ≤ min(n·m, 50)) — the dimensions of the chocolate bar and the number of squares you want to eat respectively.
Output
For each n, m and k print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly k squares.
Examples
input
4
2 2 1
2 2 3
2 2 2
2 2 4
output
5
5
4
0
Note
In the first query of the sample one needs to perform two breaks:
to split 2 × 2 bar into two pieces of 2 × 1 (cost is 22 = 4),
to split the resulting 2 × 1 into two 1 × 1 pieces (cost is 12 = 1).
In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.
题目大意:一块n*m的巧克力,要吃掉k块。我们通过沿横线或者竖线划分巧克力的方式来得到k块巧克力。沿横线分割巧克力的代价是巧克力的横线长的平方(
思路:用f[i][j][k]表示在i*j的巧克力块中吃掉k块的最小代价。对于每一种切的方式我们可以枚举分割点,将巧克力分成两块计算。横着切:则(分割点)1<=h<=i/2,上面一半巧克力需要kk块,则下面一半需要k-kk块,这样就可能横着切的状态转移方程:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int f[35][35][55];int n,m,k;void dp(){ memset(f,0x3f,sizeof(f)); for(int i=0;i<=30;i++) for(int j=0;j<=30;j++) { f[i][j][0]=0; if(i*j<=50) f[i][j][i*j]=0; } for(int i=1;i<=30;i++) for(int j=1;j<=30;j++) { int tot=min(i*j,50); for(int k=1;k<=tot;k++)//总共要切的份数 { for(int h=1;h<=i/2;h++)//横着切 { int tot1=min(h*j,k); for(int kk=0;kk<=tot1;kk++) f[i][j][k]=min(f[i][j][k],f[h][j][kk]+f[i-h][j][k-kk]+j*j); } for(int v=1;v<=j/2;v++)//竖着切 { int tot2=min(v*i,k); for(int kk=0;kk<=tot2;kk++) f[i][j][k]=min(f[i][j][k],f[i][v][kk]+f[i][j-v][k-kk]+i*i); } } }}int main(){ dp(); int T; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&k); printf("%d\n",f[n][m][k]); } return 0;}
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