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Both Saya and Kudo like balloons. One day, they heard that in the central park,
there will be thousands of people fly balloons to pattern a big image.
They were very interested about this event, and also curious about the image.
Since there are too many balloons, it is very hard for them to compute anything they need. Can you help them?
You can assume that the image is an N*N matrix, while each element can be either balloons or blank.
Suppose element A and element B are both balloons. They are connected if:
i) They are adjacent;
ii) There is a list of element C1, C2,
… , Cn,
while A and C1 are
connected, C1 and C2 are
connected …Cn and B are
connected.
And a connected block means that every pair of elements in the block is connected, while any element in the block is not connected with any element out of the block.
To Saya, element A(xa,ya)and B(xb,yb) is adjacent if |xa-xb| + |ya-yb| ≤ 1
But to Kudo, element A(xa,ya) and
element B (xb,yb) is adjacent if |xa-xb|≤1 and |ya-yb|≤1
They want to know that there’s how many connected blocks with there own definition of adjacent?
5 11001 00100 11111 11010 10010 0
Case 1: 3 2
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
struct node
{
int x;
int y;
};
int jx1[] = {1,0,0,-1};
int jy1[] = {0,1,-1,0};
int jx2[] = {1,0,-1,0,-1,1,1,-1};
int jy2[] = {0,1,0,-1,-1,1,-1,1};
char map[110][110];
int v[110][110];
int n;
void BFS1(int x,int y)
{
struct node t,f;
queue<node>q1;
t.x = x;
t.y = y;
q1.push(t);
v[t.x][t.y] = 1;
while(!q1.empty())
{
t = q1.front();
q1.pop();
for(int i=0; i<4; i++)
{
f.x = t.x + jx1[i];
f.y = t.y + jy1[i];
if(f.x>=0 && f.x<n && f.y>=0 && f.y<n && v[f.x][f.y] == 0 && map[f.x][f.y] == '1')
{
q1.push(f);
v[f.x][f.y] = 1;
}
}
}
}
void BFS2(int x,int y)
{
struct node t,f;
queue<node>q2;
t.x = x;
t.y = y;
q2.push(t);
v[t.x][t.y] = 1;
while(!q2.empty())
{
t = q2.front();
q2.pop();
for(int i=0; i<8; i++)
{
f.x = t.x + jx2[i];
f.y = t.y + jy2[i];
if(f.x>=0 && f.x<n && f.y>=0 && f.y<n && v[f.x][f.y] == 0 && map[f.x][f.y] == '1')
{
q2.push(f);
v[f.x][f.y] = 1;
}
}
}
}
int main()
{
int kk = 0;
while(scanf("%d",&n)!=EOF)
{
if(n == 0)
{
break;
}
memset(map,0,sizeof(map));
for(int i=0; i<n; i++)
{
scanf("%s",map[i]);
}
memset(v,0,sizeof(v));
int ans = 0;
int cnt = 0;
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(v[i][j] == 0)
{
if(map[i][j] == '1')
{
BFS1(i,j);
ans++;
}
}
}
}
memset(v,0,sizeof(v));
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
if(v[i][j] == 0)
{
if(map[i][j] == '1')
{
BFS2(i,j);
cnt++;
}
}
}
}
printf("Case %d: %d %d\n",++kk,ans,cnt);
printf("\n");
}
return 0;
}
SDUT 2152 Balloons（BFS 第一届山东省赛）
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原文地址：http://blog.csdn.net/yeguxin/article/details/45315547
