poj2386 poj3620
来源:互联网 发布:exm网络意思 编辑:程序博客网 时间:2024/06/10 12:09
Description
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* 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
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
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
#include <stdio.h>#include<memory.h>char a[111][111];int m,n,f;void dfs(int i,int j){ if(a[i][j]=='.'||i<0||j<0||i>m-1||j>n-1) return ; a[i][j]='.'; dfs(i,j+1); dfs(i,j-1); dfs(i+1,j); dfs(i+1,j-1); dfs(i+1,j+1); dfs(i-1,j); dfs(i-1,j+1); dfs(i-1,j-1);}int main(){ scanf("%d%d",&m,&n); f=0; memset(a,0,sizeof(a)); for(int i=0;i<m;i++) scanf("%s",a[i]); for(int i=0;i<m;i++) for(int j=0;j<n;j++) { if(a[i][j]=='W') { f++; dfs(i,j); } } printf("%d\n",f); return 0;}
Description
Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the largest "lake" on his farm.
The farm is represented as a rectangular grid with N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactlyK (1 ≤ K ≤ N × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected cell and is part of the lake.
Input
* Line 1: Three space-separated integers: N, M, and K
* Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column: R and C
Output
* Line 1: The number of cells that the largest lake contains.
Sample Input
3 4 53 22 23 12 31 1
Sample Output
4
Source
#include <cstdio>#include <iostream>#include <memory.h>using namespace std;int a[110][110],vis[110][110];int n,m,k;int dfs(int i,int j){ if(i<1||j<1||i>n||j>m||vis[i][j]!=0||a[i][j]==0) return 0; vis[i][j]=1; int t=1; t+=dfs(i-1,j); t+=dfs(i+1,j); t+=dfs(i,j+1); t+=dfs(i,j-1); return t;}int main(){ while(scanf("%d%d%d",&n,&m,&k)!=EOF) { int i,j,p,q,max; memset(a,0,sizeof(a)); memset(vis,0,sizeof(vis)); max=0; for(i=0;i<k;i++) { scanf("%d%d",&p,&q); a[p][q]=1; } for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(a[i][j]&&!vis[i][j]) { int sum=dfs(i,j); if(sum>max) max=sum; } printf("%d\n",max); } return 0;}
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