CodeForces：699B-One Bomb

B. One Bomb

time limit per test1 second

memory limit per test256 megabytes

Problem Description

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (“.”) or it can be occupied by a wall (“*”).

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

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

The next n lines contain m symbols “.” and “” each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to “.”, then the corresponding cell is empty, otherwise it equals “” and the corresponding cell is occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print “NO” in the first line (without quotes).

Otherwise print “YES” (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

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就是一个大水题啊，当时看了半天居然没有反应。

可以先算出每一行，每一列的’*’的总和放在r[],c[]数组中，然后再枚举每一点为放炸弹的中心，判断当前r[i]+c[j]（注意是否减一），是否是总和。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1100;
char maps[maxn][maxn];
int r[maxn],c[maxn];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%s",maps[i]+1);
        int sum = 0;
        for(int i=1;i<=n;i++)
        {
            int temp = 0;
            for(int j=1;j<=m;j++)
            {
                if(maps[i][j] == '*')
                    temp++;
            }
            r[i] = temp;
            sum += temp;
        }
        for(int j=1;j<=m;j++)
        {
            int temp = 0;
            for(int i=1;i<=n;i++)
            {
                if(maps[i][j] == '*')
                    temp++;
            }
            c[j] = temp;
        }

        bool flag = false;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                int temp;
                if(maps[i][j] == '*')//如果当前点是‘*’要减1
                    temp = r[i] + c[j] - 1;
                else
                    temp = r[i] + c[j];
                if(temp == sum)
                {
                    printf("YES\n%d %d\n",i,j);
                    flag = true;
                    break;
                }
            }
            if(flag)
                break;
        }
        if(!flag)
            printf("NO\n");
    }
}