toj 4612 A Shooting Game

时间限制(普通/Java):1000MS/3000MS 内存限制:65536KByte
总提交: 4 测试通过:4

描述

A and B are playing a shooting game on a battlefield consisting of square-shaped unit blocks. The blocks are occupying some consecutive columns, and the perimeter of the figure equals the perimeter of its minimal bounding box. The figure (a) below is a valid battlefield, but (b) and (c) are not, because in (b), there is an empty column; in (c), the perimeter of figure is 14, but the perimeter of the bounding box (drawn with dashed lines) is 12. With the help of gravity, each block is either located on another block, or sitting on the ground. To make the battlefield look more exciting, it must not be a perfect rectangle (i.e. it is not allowed that every column has the same height)

Here is the rule of the game:

A and B shoot by turn. A starts first.

Before shooting, the player first select one row with at least one block, and one of the two directions “left” and “right”, then shoot at this row along that direction. The power of the shoot is one of 1, 2 or 3, each with probability of 1/3. The power of shoot is the number of blocks (not necessarily consecutive) that can be destroyed in this shoot. If the total number of blocks on this row is less than the power of shoot, then all the blocks on this row is destroyed. For example, if the player chooses to shoot the 3rd row from the bottom, with direction “right”, power 2, and there are 4 blocks on this row, then the left-most two blocks are destroyed.

After each shoot, blocks in the air fall down vertically. The next player cannot shoot before all the blocks stop falling.

Realize that the intermediate battlefields do not have to follow the constraints for starting battlefields. For example, it could happens some situations looking as figures (b) or (c), and then, if the power is p, the leftmost/rightmost p blocks of columns which contain a block in this row are destroyed (skipping empty positions).

He who destroys the last block wins.

Assume the starting battlefield is . According to rule 1, A shoots first. The table below shows three (not all) possible outcomes of the first shot:

Assume Alice and Bob are both very clever (always follows the strategy that maximizes the probability he/she wins), what is the probability that Alice wins?

输入

There will be at most 25 test cases, each with two lines. The first line is a single integer n ( 1n6), the number of columns. The second line contains nintegers h1, h2, …, hn ( 1hi6), the heights of the columns from left to right. The battlefield is guaranteed to satisfy the restrictions in the problem (perimeter of figure equals that of the minimal bounding box, and is not a perfect rectangle). Input is terminated by n = 0.

输出

For each test case, print a single line, the probability that A wins, to six decimal points.

样例输入

3
2 1 1

样例输出

0.555556

/*
本题采用记忆化搜索*/#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
#define FOR(i, s, t) for(int i = s; i <= t; ++i)const int MAXS = 50010;
const int MAXN = 8;
const double EPS = 1e-8;double a[7][7][7][7][7][7];double solve(int p1, int p2, int p3, int p4, int p5, int p6) {//solve函数求当前状态下获胜的概率的最大值if(a[p1][p2][p3][p4][p5][p6] > EPS) return a[p1][p2][p3][p4][p5][p6];//若当前位置已经存储了此状态下的最优解，则直接使用if(p1 + p2 + p3 + p4 + p5 + p6 == 0) return 0;//所有的列都为0，表示所有的木块全部打完int v1, v2, v3, v4, v5, v6;double ret = 0;//记录最优解for(int i = 1; i <= 6; ++i) {//从左边打的情况double tmp = 0;//i从1到6各个情况下的解for(int j = 1; j <= 3; ++j) {//射击威力的随机值v1 = p1, v2 = p2, v3 = p3, v4 = p4, v5 = p5, v6 = p6;//将原始数据另存，保证每次都从此状态出发搜索，每一行木块的个数int k = j;//记录当前的威力if(v1 >= i && k) --v1, --k;//if(v2 >= i && k) --v2, --k;if(v3 >= i && k) --v3, --k;if(v4 >= i && k) --v4, --k;if(v5 >= i && k) --v5, --k;if(v6 >= i && k) --v6, --k;if(k == j) continue;//没打中，此轮轮空tmp += 1./3 * (1 - solve(v1, v2, v3, v4, v5, v6));//每次循环的结果都是1/3，乘以另一个人射击赢的概率，所以用反面考虑}if(tmp > ret) ret = tmp;//如果概率大于记录值则更新记录值tmp = 0;for(int j = 1; j <= 3; ++j) {//左边情况同上v1 = p1, v2 = p2, v3 = p3, v4 = p4, v5 = p5, v6 = p6;int k = j;if(v6 >= i && k) --v6, --k;if(v5 >= i && k) --v5, --k;if(v4 >= i && k) --v4, --k;if(v3 >= i && k) --v3, --k;if(v2 >= i && k) --v2, --k;if(v1 >= i && k) --v1, --k;if(k == j) continue;tmp += 1./3 * (1 - solve(v1, v2, v3, v4, v5, v6));}if(tmp > ret) ret = tmp;}return a[p1][p2][p3][p4][p5][p6] = ret;//记录当前步最优解
}int x[7], n;int main() {//FOR(i1, 0, 6) FOR(i2, 0, 6) FOR(i3, 0, 6) FOR(i4, 0, 6) FOR(i5, 0, 6) FOR(i6, 0, 6)//if(a[i1][i2][i3][i4][i5][i6] < EPS) solve(i1, i2, i3, i4, i5, i6);while(scanf("%d", &n) != EOF && n) {memset(x, 0, sizeof(x));for(int i = 1; i <= n; ++i) scanf("%d", &x[i]);printf("%.6f\n", solve(x[1], x[2], x[3], x[4], x[5], x[6]));}
}