CFhM_R's Blog

题目： 代码1： #include #include #include #include using namespace std; int map[1010][1010]; int dx[] = { 0,0,1,-1 }; int dy[] = { 1,-1,0,0 }; int n, m,ans,x2,y2; //void init() //{ // memse

题目： Curling 2.0 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16112   Accepted: 6657 Description On Planet MM-21, after their Olympic games this yea

题目： Beat Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1203    Accepted Submission(s): 716 Problem Description Zty is a man th

题目： Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 6775   Accepted: 2958   Special Judge Description Bessie's been appointed the new watch-c

题目： Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 98266    Accepted Submission(s): 26642 Problem Description

题目： Zipper Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8645    Accepted Submission(s): 3061 Problem Description Given three

【poj3321】Apple Tree——树状数组&DFS##include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 100005; struct edge { int v,nxt; }e[maxn]; int head[maxn],c[maxn],low[m