传送门 DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 52 Problem Description A topological sort or topological ordering of a directed…

DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 866    Accepted Submission(s): 250 Problem Description A topological sort or topological ordering of a directed g…

DZY Loves Topological Sorting Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5195 Description A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for ev…

DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1250    Accepted Submission(s): 403 Problem Description A topological sort or topological ordering of a directed g…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5195 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=573&pid=1002 题解: 1.拓扑排序+贪心 #include<algorithm> #include<iostream> #include<cstring> #include<cstdi…

题意: 删去K条边,使拓扑排序后序列字典序最大 分析: 因为我们要求最后的拓扑序列字典序最大,所以一定要贪心地将标号越大的点越早入队.我们定义点i的入度为di. 假设当前还能删去k条边,那么我们一定会把当前还没入队的di≤k的最大的i找出来,把它的di条入边都删掉,然后加入拓扑序列. 删除的一定是小连大的边,因为大连小的边在拓扑序列生成的时候就去掉了 #include <iostream> #include <cstdio> #include <queue> #incl…

题目链接:pid=5228">ZCC loves straight flush pid=5228">题面: pid=5228"> ZCC loves straight flush Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 827    Accepted Submission(s): 340…

若 [i, j] 满足, 则 [i, j+1], [i, j+2]...[i,n]均满足 故设当前区间里个数为size, 对于每个 i ,找到刚满足 size == k 的 [i, j], ans += n - j + 1 . i++ 的时候看看需不需要size-- 就可以更新了. #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define LL long l…

先找相邻差值的最大,第二大,第三大 删去端点会减少一个值, 删去其余点会减少两个值,新增一个值,所以新增和现存的最大的值比较一下取最大即可 #include <iostream> #include <cstdio> #include <cmath> using namespace std; #define LL long long ; int t, n, p1, p2, p3; LL a[N]; LL s1[N], s2[N]; LL sum; int main() {…

传送门 题意简述:给出一张DAGDAGDAG,要求删去不超过kkk条边问最后拓扑序的最大字典序是多少. 思路:贪心帮当前不超过删边上限且权值最大的点删边,用线段树维护一下每个点的入度来支持查询即可. 注意要在选点的时候更新后继的入度 代码: #include<bits/stdc++.h> #define ri register int using namespace std; inline int read(){ int ans=0; char ch=getchar(); while(!isd…