• hdoj-1312-Red and Black


    Red and Black

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 12433 Accepted Submission(s): 7726


    Problem Description
    There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

    Write a program to count the number of black tiles which he can reach by repeating the moves described above.

    Input
    The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

    There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

    '.' - a black tile
    '#' - a red tile
    '@' - a man on a black tile(appears exactly once in a data set)

    Output
    For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

    Sample Input
    6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0

    Sample Output
    45 59 6 13

    Source

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    #include<stdio.h>
    #include<string.h>
    char map[22][22];
    int  dd[22][22];
    int  n,m;
    void DP(int i,int j,int &res){
    	if(i<=0||i>n||j<0||j>=m) return ;
    	if(map[i][j]=='#') return;
    	if(dd[i][j]==0&&(map[i][j]=='.'||map[i][j]=='@')) {
    		dd[i][j]=1;
    		res++;
    		DP(i-1,j,res);DP(i+1,j,res);
    		DP(i,j-1,res);DP(i,j+1,res);
    	}
    }
    int main(){
    	while(scanf("%d %d",&m,&n),n||m){
    		memset(map,0,sizeof(map));
    		memset(dd,0,sizeof(dd));
    		int i,j,res,num=0;int tx,ty;
    
    		for(i=1;i<=n;++i){
    	            scanf("%s",map[i]);		
    		}
    		for(i=1;i<=n;++i){
    			for(j=0;j<m;++j)
    			  if(map[i][j]=='@') 
    			    tx=i,ty=j;
    		}
    		//printf("%d====%d
    ",tx,ty);;
    		res=0;
    		DP(tx,ty,res);
    		printf("%d
    ",res);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/mthoutai/p/7063982.html
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