BFS求连通块.递归会爆栈. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<map> #include<queue> #include<stack> #include<vector> using namespace std; int M,N,L,T; ][][]; ][][]; ][]={ {,,}…

题意: 输入四个正整数M,N,K,T(K<=60,M<=1286,N<=128),代表每片的高度和宽度,片数和最小联通块大小.输出一共有多少个单元满足所在联通块大小大于等于T. trick: 三元数组大小开小了...最后两个测试点答案错误,我是笨比. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; int m,n,l,t; ][][]; ]…

题目如下: One important factor to identify acute stroke (急性脑卒中) is the volume of the stroke core. Given the results of image analysis in which the core regions are identified in each MRI slice, your job is to calculate the volume of the stroke core. Inpu…

One important factor to identify acute stroke (急性脑卒中) is the volume of the stroke core. Given the results of image analysis in which the core regions are identified in each MRI slice, your job is to calculate the volume of the stroke core. Input Spec…

题意:给定三维数组,0表示正常,1表示有肿瘤块,肿瘤块的区域>=t才算是肿瘤,求所有肿瘤块的体积和 这道题一开始就想到了dfs或者bfs,但当时看数据量挺大的,以为会导致栈溢出,所以并没有立刻写,想有没有别的办法.然而结果是,实在想不出别的办法了,所以还是尝试写写dfs.bfs. 一开始先用了dfs,最后两个样例段错误,估计是栈溢出了.之所以dfs栈溢出,因为dfs的时候每个状态都会存储在堆栈里,就好比dfs的第一个状态,一直保存到最后整个dfs结束.而bfs是存储在队列中,每次队列都会有状态取…

预处理出最短路再进行暴力dfs求答案会比较好.直接dfs效率太低. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; int n,m; int st,en…

简单并查集. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; +; int fa[maxn]; int num[maxn]; int n; vector…

暴力搜索. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; int n,k,p,top; ]; ; ],ans_num=,ans_len; void d…

枚举一下选的位置,每次算一下就可以了. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; const int…

数位DP.dp[i][j]表示i位,最高位为j的情况下总共有多少1. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namesp…