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  • cf #363 b

    B. One Bomb
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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.

    Examples
    input
    3 4
    .*..
    ....
    .*..
    output
    YES
    1 2
    input
    3 3
    ..*
    .*.
    *..
    output
    NO
    input
    6 5
    ..*..
    ..*..
    *****
    ..*..
    ..*..
    ..*..
    output
    YES
    3 3



    #include<stdio.h>
    #include<iostream>
    #include<map>
    using namespace std;
    char ma[1005][1005];
    int main(){
        int n,m;
        int x=-1;
        int y=-1;
        int hang[1005];
        int lie[1005];
        int cnt=0;
        scanf("%d%d",&m,&n);
        for(int i=0;i<m;i++) hang[i]=0;
        for(int j=0;j<n;j++) lie[j]=0;
        for(int i=0;i<m;i++){
            scanf("%s",ma[i]);
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(ma[i][j]=='*'){
                    hang[i]++;
                    lie[j]++;
                    cnt++;
                }
            }
        
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                int tmp=hang[i]+lie[j];
                if(ma[i][j]=='*') tmp--;
                if(tmp==cnt){
                    printf("YES
    %d %d",i+1,j+1);
                    return 0;
                }
            }
        }
        printf("NO
    ");
        return 0;
    }
    View Code

    分别统计行和列中到墙的个数,如果某个 行和列中包含全部到墙,则为答案。

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  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5690607.html
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