LightOJ1284 Lights inside 3D Grid (概率DP)
You are given a 3D grid, which has dimensions X, Y and Z. Each of the X x Y x Z cells contains a light. Initially all lights are off. You will have K turns. In each of the K turns,
- You select a cell A randomly from the grid,
- You select a cell B randomly from the grid and
- Toggle the states of all the bulbs bounded by cell A and cell B, i.e. make all the ON lights OFF and make all the OFF lights ON which are bounded by A and B. To be clear, consider cell A is (x1, y1, z1) and cell B is (x2, y2, z2). Then you have to toggle all the bulbs in grid cell (x, y, z) where min(x1, x2) ≤ x ≤ max(x1, x2), min(y1, y2) ≤ y ≤ max(y1, y2) and min(z1, z2) ≤ z ≤ max(z1, z2).
Your task is to find the expected number of lights to be ON after K turns.
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
Each case starts with a line containing four integers X, Y, Z (1 ≤ X, Y, Z ≤ 100) and K (0 ≤ K ≤ 10000).
Output
For each case, print the case number and the expected number of lights that are ON after K turns. Errors less than 10-6will be ignored.
Sample Input
5
1 2 3 5
1 1 1 1
1 2 3 0
2 3 4 1
2 3 4 2
Sample Output
Case 1: 2.9998713992
Case 2: 1
Case 3: 0
Case 4: 6.375
Case 5: 9.09765625
题解:
参考代码:
#include<bits/stdc++.h> using namespace std; typedef long long ll; int T,X,Y,Z,K; double ans; double calc(int x,int m) { return 1.0-1.0*((m-x)*(m-x)+(x-1)*(x-1))/(m*m); } int main() { scanf("%d",&T); for(int cas=1;cas<=T;++cas) { ans=0; scanf("%d%d%d%d",&X,&Y,&Z,&K); for(int i=1;i<=X;++i) for(int j=1;j<=Y;++j) for(int k=1;k<=Z;++k) { double p=calc(i,X)*calc(j,Y)*calc(k,Z); ans+=0.5-0.5*pow(1.0-2*p,K); } printf("Case %d: %.8lf\n",cas,ans); } return 0; }