哈理工2015暑假集训 zoj 2975 Kinds of Fuwas

最新推荐文章于 2018-07-19 00:02:25 发布

转载最新推荐文章于 2018-07-19 00:02:25 发布 · 43 阅读

本文介绍了一个基于福娃矩阵的游戏算法，玩家需要找出所有四个角相同种类福娃构成的矩形，文章提供了一种解决方案，利用两次嵌套遍历列来统计相同福娃的数量并计算矩形数量。

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G - Kinds of Fuwas

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2975

Description

In the year 2008, the 29th Olympic Games will be held in Beijing. This will signify the prosperity of China as well as becoming a festival for people all over the world.

The official mascots of Beijing 2008 Olympic Games are Fuwa, which are named as Beibei, Jingjing, Haunhuan, Yingying and Nini. Fuwa embodies the natural characteristics of the four most popular animals in China -- Fish, Panda, Tibetan Antelope, Swallow -- and the Olympic Flame. To popularize the official mascots of Beijing 2008 Olympic Games, some volunteers make a PC game with Fuwa.

As shown in the picture, the game has a matrix of Fuwa. The player is to find out all the rectangles whose four corners have the same kind of Fuwa. You should make a program to help the player calculate how many such rectangles exist in the Fuwa matrix.

Input

Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <= T <= 50) which is the number of test cases. And it will be followed byT consecutive test cases.

The first line of each test case has two integers M and N (1 <=M, N <= 250), which means the number of rows and columns of the Fuwa matrix. And then there areM lines, each has N characters, denote the matrix. The characters -- 'B' 'J' 'H' 'Y' 'N' -- each denotes one kind of Fuwa.

Output

Results should be directed to standard output. The output of each test case should be a single integer in one line, which is the number of the rectangles whose four corners have the same kind of Fuwa.

Sample Input

2
2 2
BB
BB
5 6
BJHYNB
BHBYYH
BNBYNN
JNBYNN
BHBYYH

Sample Output

1
8

当时训练赛的时候做的也比較麻烦，并没有非常成熟的思维去迅速组织搞笑代码。这是改进后的代码。

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
const int INF=260;
char cnt[INF][INF];
int n,m;

void before()
{
    cin>>n>>m;
    for(int i=0; i<n; i++)
        scanf("%s",cnt[i]);
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        before();
        int sum=0;
        for(int i=0; i<m-1; i++)//前两个for循环 控制列变换
        {
            for(int j=i+1; j<m; j++)
            {              
                map< char ,int >dict;    //  统计每一个字母个数
                set<char>xx;           //  统计字母种类
                for(int k=0; k<n; k++) //控制行数
                {
                    if(cnt[k][i]==cnt[k][j])
                    {
                        dict[cnt[k][j]]++;
                        xx.insert(cnt[k][j]);
                    }
                }
                set<char >::iterator it;
                for(it=xx.begin(); it!=xx.end(); it++)
                {
                    //如果某两列 出现n对同样的 则矩形数目为 n*(n-1)/2个
                    sum+=dict[*it]*(dict[*it]-1)/2;
                }
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}