PAT甲级1091 - Acute Stroke
2018-02-26 17:32
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相当于一个连通图,只不过方向由二维的四个方向改成了三维上的6个方向。找到每个连通子图,判断是否大于或等于t。最后计算总和即可。
这个不能用dfs递归的进行,会出现段错误。改成bfs,可以AC。
这个不能用dfs递归的进行,会出现段错误。改成bfs,可以AC。
#include<bits/stdc++.h> using namespace std; int m,n,l,t; bool in[60+4][1286+4][128+4],visited[60+4][1286+4][128+4]; typedef struct node{ int x,y,z; node(){} node(int x,int y,int z){ this->x=x; this->y=y; this->z=z; } }point; void check(queue<point> &p,point a,int dx,int dy,int dz){ int x=a.x,y=a.y,z=a.z; if(!visited[x+dx][y+dy][z+dz] && in[x+dx][y+dy][z+dz]){ visited[x+dx][y+dy][z+dz]=true; p.push(point(x+dx,y+dy,z+dz)); //visited[x+dx][y+dy][z+dz]=false; } } int bfs(point a){ int cnt=0; queue<point> p; p.push(a); point b; while(!p.empty()){ ++cnt; b= p.front();p.pop(); check(p,b,0,-1,0); check(p,b,0,1,0); check(p,b,0,0,1); check(p,b,0,0,-1); check(p,b,-1,0,0); check(p,b,1,0,0); } return cnt; } int main(){ scanf("%d %d %d %d",&m,&n,&l,&t); for(int i=1;i<=l;++i){ for(int j=1;j<=m;++j){ for(int k=1;k<=n;++k){ scanf("%d",&in[i][j][k]); } } } int ans=0,cnt; for(int i=1;i<=l;++i){ for(int j=1;j<=m;++j){ for(int k=1;k<=n;++k){ if(in[i][j][k] && !visited[i][j][k]){ visited[i][j][k]=true; cnt = bfs(point(i,j,k)); if(cnt>=t){ ans+=cnt; } } } } } printf("%d",ans); return 0; }
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