UVALive 4622 Decision（bfs）

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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;}

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