Codeforces_738B

最新推荐文章于 2018-04-11 10:32:59 发布

转载最新推荐文章于 2018-04-11 10:32:59 发布 · 69 阅读

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原文链接：http://www.cnblogs.com/jasonlixuetao/p/6085754.html

本文解析了一道关于在给定的剧场舞台上寻找放置聚光灯的良好位置的算法题目。良好位置需满足两个条件：放置位置无演员且投影方向至少有一个演员。文章提供了完整的C++代码实现，并通过示例详细解释了如何计算良好位置的数量。

Theater stage is a rectangular field of size n × m. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.

You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.

A position is good if two conditions hold:

there is no actor in the cell the spotlight is placed to;
there is at least one actor in the direction the spotlight projects.

Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the plan.

The next n lines contain m integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.

Output

Print one integer — the number of good positions for placing the spotlight.

Examples

input

2 4
0 1 0 0
1 0 1 0

output

input

output

Note

In the first example the following positions are good:

the (1, 1) cell and right direction;
the (1, 1) cell and down direction;
the (1, 3) cell and left direction;
the (1, 3) cell and down direction;
the (1, 4) cell and left direction;
the (2, 2) cell and left direction;
the (2, 2) cell and up direction;
the (2, 2) and right direction;
the (2, 4) cell and left direction.

Therefore, there are 9 good positions in this example.

一道水题，比赛时没看复杂度，终测时TLE了，醉人。。。

#include<iostream>
#include<cstdio>
using namespace std;
#define N 1005

int gra[N][N];
int main()
{
        int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            scanf("%d",&gra[i][j]);
    int res=0;
    for(int i=0; i<n; i++)
    {
        int st=-1,en=-1;
        for(int j=0; j<m; j++)
        {
            if(gra[i][j]==1)
            {
                if(st==-1)
                    st=j;
                en=j;
            }
        }
        if(st>=0)
        {
            res+=st;
            for(int k=0;k<st;k++)
                if(gra[i][k]==1)
                        res--;
            if(en>st)
            {
                    res+=(en-st-1)*2;
                    for(int k=st+1;k<en;k++)
                        if(gra[i][k]==1)
                                res-=2;
            }
            res+=(m-1-en);
            for(int k=en+1;k<m;k++)
                if(gra[i][k]==1)
                        res--;
        }
    }

    for(int j=0; j<m; j++)
    {
        int st=-1,en=-1;
        for(int i=0; i<n; i++)
        {
            if(gra[i][j]==1)
            {
                if(st==-1)
                    st=i;
                en=i;
            }
        }
        if(st>=0)
        {
            res+=st;
            for(int k=0;k<st;k++)
                if(gra[k][j]==1)
                        res--;
             if(en>st)
            {
                    res+=(en-st-1)*2;
                    for(int k=st+1;k<en;k++)
                        if(gra[k][j]==1)
                                res-=2;
            }
            res+=(n-1-en);
            for(int k=en+1;k<n;k++)
                if(gra[k][j]==1)
                        res--;
        }
    }
    printf("%d\n",res);
    return 0;
}

转载于:https://www.cnblogs.com/jasonlixuetao/p/6085754.html