A - Lake Counting解题报告
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A - Lake Counting
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uDescription
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
这道题由于学长讲过原理,所以我就照着学长讲的做了。第一次做DFS,感觉还不错嘛。
期间由于宽高搞错互换,还有读数据的时候读入了回车导致数据不正确,这种问题还是蛮多的。
原理就是检索水洼,如果这个水洼没有被DFS过的话,水洼数加一,对它DFS,把周围的水洼都标记了,这样继续检索的时候就不会在计算入内了。
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
这道题由于学长讲过原理,所以我就照着学长讲的做了。第一次做DFS,感觉还不错嘛。
期间由于宽高搞错互换,还有读数据的时候读入了回车导致数据不正确,这种问题还是蛮多的。
原理就是检索水洼,如果这个水洼没有被DFS过的话,水洼数加一,对它DFS,把周围的水洼都标记了,这样继续检索的时候就不会在计算入内了。
//Lake Counting#include <stdio.h>char Map[101][101];int MapVisited[101][101] = {0};int MapWidth,MapHeight;int WaterNum;void DFS(int x,int y);char JudgeWater(int x,int y);void main(){int i,j;//Read Map Datascanf("%d %d",&MapHeight,&MapWidth);getchar();for (j = 1;j <= MapHeight;j ++){for (i = 1;i <= MapWidth;i ++){scanf("%c",&Map[i][j]);}getchar();}/*for (j = 1;j <= MapHeight;j ++){for (i = 1;i <= MapWidth;i ++){printf("%c",Map[i][j]);}putchar('\n');}*/WaterNum = 0;//DFSfor (j = 1;j <= MapHeight;j ++){for (i = 1;i <= MapWidth;i ++){if (Map[i][j] == 'W' && MapVisited[i][j] == 0)//Not Searched Water{//printf("x %d y %d\n",i,j);WaterNum ++;DFS(i,j);}}}printf("%d\n",WaterNum);}void DFS(int x,int y){if (MapVisited[x][y] == 0){MapVisited[x][y] = 1;//printf("Vistied x %d y %d\n",x,y);if (JudgeWater(x - 1,y - 1) == 'W'){DFS(x - 1,y - 1);}if (JudgeWater(x - 1,y) == 'W'){DFS(x - 1,y);}if (JudgeWater(x - 1,y + 1) == 'W'){DFS(x - 1,y + 1);}if (JudgeWater(x,y - 1) == 'W'){DFS(x,y - 1);}if (JudgeWater(x,y + 1) == 'W'){DFS(x,y + 1);}if (JudgeWater(x + 1,y - 1) == 'W'){DFS(x + 1,y - 1);}if (JudgeWater(x + 1,y) == 'W'){DFS(x + 1,y);}if (JudgeWater(x + 1,y + 1) == 'W'){DFS(x + 1,y + 1);}}}char JudgeWater(int x,int y){if (x >= 1 && x <= MapWidth && y >= 1 && y <= MapHeight){return Map[x][y];}return '.';}
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