徐州网络赛C-Cacti Lottery【DFS】

• 54.19%
• 2000ms
• 262144K

Morgana is playing a game called cacti lottery. In this game, morgana has a 3 \times 33×3 grid graph, and the graph is filled with 11 ~ 99 , each number appears only once. The game is interesting, he doesn't know some numbers, but there are also some numbers he knows but don't want to tell you.

Now he should choose three grids, these three grids should be in the same column or in the same row or in the diagonal line. Only after this choice, can he know all numbers in the grid graph. Then he sums the three numbers in the grids he chooses, get the reward as follows:

Sum	Reward
6	10000
7	36
8	720
9	360
10	80
11	252
12	108
13	72
14	54
15	180
16	72
17	180
18	119
19	36
20	360
21	1080
22	144
23	1800
24	3600

Then he wants you to predict the expected reward he will get if he is clever enough in the condition that he doesn't want to tell you something he knows.

("He is clever enough" means he will choose the max expected reward row or column or dianonal in the condition that he knows some numbers. And you know that he knows some number, but you don't know what they are exactly. So you should predict his expected reward in your opinion. )

Input

First line contains one integers TT (T \le 100T≤100) represents the number of test cases.

Then each cases contains three lines, giving the 3 \times 33×3 grid graph. '*' means Morgana knows but doesn't want to tell you, '#' means Morgana doesn't know, '0' ~ '9' means the public number that Morgana and you both know.

Output

TT lines, output the answer. If the answer is AA, and your answer is BB. Your answer will be considered as correct if and only if |(A-B)| < 1e-5∣(A−B)∣<1e−5 .

本题答案不唯一，符合要求的答案均正确

样例输入复制

2
123
***
###
12*
45#
78#

样例输出复制

10000
4313.16666667

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

感觉这道题的难度在于题意。读第一次的时候完全不知道在干什么。

* #有什么区别也搞不懂 样例的答案凑半天凑不出来

后来Leo来说的时候突然豁然开朗

比赛后来来不及敲了 其实也不太难

因为9的阶乘是很小的 所以可以枚举每一种可能

枚举所有*的可能 算每一种的期望

其中又要枚举每一种#的可能 要把每一种#的分数都加在一起找到最大的那个 作为当前*状态的分值除以#全排列数作为期望

把所有*的期望相加 除以A(cntjin+cntxing, cntxing） 就是答案

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x7f7f7f7f7f7f7f7f
using namespace std;
typedef long long LL;
int t;
bool vis[10];
int orinum[10], num[10], ans[10];
double marks[30] = {0, 0, 0, 0, 0, 0, 10000, 36, 720, 360, 80,
252, 108, 72, 54, 180, 72, 180, 119, 36, 360, 1080, 144, 1800, 3600};
vector<int> notused;
vector<int> xingpos;
vector<int> jinpos;
double A(int n, int m)
{
    double res = 1.0;
    for(int i = 0; i < m; i++){
        res *= (n - i);
    }
    return res;
}
void maxexp()
{
    ans[1] += marks[num[0] + num[1] + num[2]];
    ans[2] += marks[num[3] + num[4] + num[5]];
    ans[3] += marks[num[6] + num[7] + num[8]];
    ans[4] += marks[num[0] + num[3] + num[6]];
    ans[5] += marks[num[1] + num[4] + num[7]];
    ans[6] += marks[num[2] + num[5] + num[8]];
    ans[7] += marks[num[0] + num[4] + num[8]];
    ans[8] += marks[num[2] + num[4] + num[6]];
}
void dfsjin(int id)
{
    //double res = 0;
    if(id == jinpos.size()){
        /*for(int i = 0; i < 9; i++){
            cout<<num[i];
        }
        cout<<" "<<maxexp()<<endl;*/
        maxexp();
    }
    else{
        for(int i = 0; i < notused.size(); i++){
            int j = notused[i];
            if(vis[j]) continue;
            num[jinpos[id]] = j;
            vis[j] = true;
            dfsjin(id + 1);
            vis[j] = false;
        }
    }
    //return res;
}
double dfs(int id)
{
    double res = 0;
    if(id == xingpos.size()){
        memset(ans, 0, sizeof(ans));
        dfsjin(0);
        int res = 0;
        for(int i = 1; i <= 9; i++){
            res = max(res, ans[i]);
        }
        return res / A(jinpos.size(), jinpos.size());
    }
    else{
        for(int i = 0; i < notused.size(); i++){
            int j = notused[i];
            if(vis[j]) continue;
            num[xingpos[id]] = j;
            vis[j] = true;
            res += dfs(id + 1);
            vis[j] = false;
        }
    }
    return res;
}
int main()
{
    cin>>t;
    while(t--){
        memset(vis, 0, sizeof(vis));
        xingpos.clear();jinpos.clear();notused.clear();
        int cntxing = 0, cntjin = 0;
        for(int i = 0; i < 3; i++){
            char s[5];
            scanf("%s", s);
            for(int j = 0; j < 3; j++){
                orinum[i * 3 + j] = num[i * 3 + j] = s[j] - '0';
                if(s[j] == '*'){
                    cntxing++;
                    xingpos.push_back(i * 3 + j);
                }
                else if(s[j] == '#'){
                    cntjin++;
                    jinpos.push_back(i * 3 + j);
                }
                else{
                    vis[num[i * 3 + j]] = true;
                }
            }
        }
        for(int i = 1; i <= 9; i++){
            if(!vis[i]){
                notused.push_back(i);
            }
        }
        double res = dfs(0) / A(cntjin + cntxing, cntxing);
        printf("%.6f\n", res);
    }
	return 0;
}