UVALive 4622 Decision（bfs）

题目链接：【UVALive 4622】

ABCDEF代表着6种不同的 区块，输入n*m的由A~F组成的字符矩阵，问这个矩阵有几个阴影区块，有公共边的是相连的

从边入手，用结构体储存四条边，有阴影的边是1，无阴影的边是0，比如B这个图像，n.l=1， n.d=1， n.r=0,，n.f=0，bfs搜索，将公共边都是1的标记

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
int t, n, m;
char str[1010][1010];
int vis[1010][1010];
int f[4][2] = {0,1,1,0,0,-1,-1,0};

struct node
{
int l, d, r, f, x, y;
}p[1010][1010], s, e;
void getp(int i, int j)
{
if(str[i][j]=='A') p[i][j].l=0, p[i][j].d=0, p[i][j].r=0, p[i][j].f=0, p[i][j].x=i, p[i][j].y=j;
else if(str[i][j]=='B') p[i][j].l=1, p[i][j].d=1, p[i][j].r=0, p[i][j].f=0, p[i][j].x=i, p[i][j].y=j;
else if(str[i][j]=='C') p[i][j].l=1, p[i][j].d=0, p[i][j].r=0, p[i][j].f=1, p[i][j].x=i, p[i][j].y=j;
else if(str[i][j]=='D') p[i][j].l=0, p[i][j].d=0, p[i][j].r=1, p[i][j].f=1, p[i][j].x=i, p[i][j].y=j;
else if(str[i][j]=='E') p[i][j].l=0, p[i][j].d=1, p[i][j].r=1, p[i][j].f=0, p[i][j].x=i, p[i][j].y=j;
else if(str[i][j]=='F') p[i][j].l=1, p[i][j].d=1, p[i][j].r=1, p[i][j].f=1, p[i][j].x=i, p[i][j].y=j;
}
void bfs(int xi, int yi, int li)
{
queue<node>q;
s = p[xi][yi];
s.x=xi, s.y=yi;
q.push(s);
while(!q.empty())
{
s = q.front(), q.pop();
if(s.l)
{
e = s;
e.y--;
e = p[e.x][e.y];
if(e.r&&!vis[e.x][e.y]&&(e.x>=1&&e.x<=n&&e.y>=1&&e.y<=m))
{
vis[e.x][e.y] = li;
q.push(e);
}
}
if(s.d)
{
e = s;
e.x++;
e = p[e.x][e.y];
if(e.f&&!vis[e.x][e.y]&&(e.x>=1&&e.x<=n&&e.y>=1&&e.y<=m))
{
vis[e.x][e.y] = li;
q.push(e);
}
}
if(s.r)
{
e = s;
e.y++;
e = p[e.x][e.y];
if(e.l&&!vis[e.x][e.y]&&(e.x>=1&&e.x<=n&&e.y>=1&&e.y<=m))
{
vis[e.x][e.y] = li;
q.push(e);
}
}
if(s.f)
{
e = s;
e.x--;
e = p[e.x][e.y];
if(e.d&&!vis[e.x][e.y]&&(e.x>=1&&e.x<=n&&e.y>=1&&e.y<=m))
{
vis[e.x][e.y] = li;
q.push(e);
}
}
}
}
int main()
{
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++)
{
scanf("%s", str[i]+1);
for(int j=1; j<=m; j++)
{
getp(i, j);
}
}
int li=0;
memset(vis, 0, sizeof(vis));
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if(vis[i][j]) continue;
vis[i][j] = 1;
if(str[i][j]=='A') continue;
li++;
bfs(i, j, li);
}
}
printf("%d\n", li);
}
return 0;
}