题意: 有N个城市,M条单向路,Tom想环游全部城市,每次至少环游2个城市,每个城市只能被环游一次.由于每条单向路都有长度,要求游遍全部城市的最小长度. // 给定一个有向图,必须用若干个环来覆盖整个图,要求这些覆盖的环的权值最小. XD 第一次知道网络流拆点做法 但还没完全理解透,多做几题感悟感悟 ,先发一下别人题解 ,代码当然自己写 思路: 原图每个点 u 拆为 u 和 u' ,从源点引容量为 1 费用为 0 的边到 u ,从 u' 引相同性质的边到汇点,若原图中存在 (u, v) ,则从 

Problem Description There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1853 Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others) Total Submission(s): 1904    Accepted Submission(s): 951 Problem Description There are N cities in our c

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 2399    Accepted Submission(s): 1231 Problem Description There are N cities in our country, and M one-way roads connecting them. Now L

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1197    Accepted Submission(s): 626 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others) Total Submission(s): 1879    Accepted Submission(s): 938 Problem Description There are N cities in our country, and M one-way roads connecting them. Now L

题目传送门 /* KM: 相比HDOJ_1533,多了重边的处理,还有完美匹配的判定方法 */ #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; int x[MAXN], y[MAXN]; int w[MAXN][MAXN]; int visx[MAXN],

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1853 Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 2289    Accepted Submission(s): 1162 Problem Description There are N cities in our

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1399    Accepted Submission(s): 712 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others) Total Submission(s): 1120    Accepted Submission(s): 579 Problem Description There are N cities in our country, and M one-way roads connecting them. Now L

Cyclic Tour                                                                                Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)                                                                            

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 2257    Accepted Submission(s): 1148 Problem Description There are N cities in our country, and M one-way roads connecting them. Now L

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1688    Accepted Submission(s): 859 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li

[线段树I:母题]hdu 1166 敌兵布阵 题目链接:hdu 1166 敌兵布阵 题目大意 C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又開始忙乎了.A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况. 因为採取了某种先进的监測手段,所以每一个工兵营地的人数C国都掌握的一清二楚,每一个工兵营地的人数都有可能发生变动.可能添加或降低若干人手,但这些都逃只是C国的监视. 中央情报局要研究敌人到底演习什么战术,所以

[题解]HDU Homework(倍增) 矩阵题一定要多多检查一下是否行列反了... 一百个递推项一定要存101个 说多了都是泪啊 一下午就做了这一道题因为实在是太菜了太久没写这种矩阵的题目... 设一个行向量\(e\),和一个增逛矩阵\(A\),他们咋定义的见我那篇讲线性递推博客 现在我们再预处理\(st\)矩阵数组,其中\(st_i=A^{2^i}\). 考虑这样一种做法,我们考虑让\(e\)总共有101个值,然后当第一个值被增逛为\(f_{q.n-100}\)时,暴力将\(e\)中的第\(

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1853 There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city b

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10685    Accepted Submission(s): 4583 Problem Description CC always becomes very depressed at the end of this month, he has checked his credit car

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5489 题目大意: 一个N(N<=100000)个数的序列,要从中去掉相邻的L个数(去掉整个区间),使得剩余的数最长上升子序列(LIS)最长. 题目思路: [二分][最长上升子序列] 首先,假设去掉[i,i+m-1]这L个数,剩余的LIS长度为max(i左端最后一个不大于a[i+m]的LIS长度+a[i+m]开始到最后的LIS长度). 所以,我们从n到1逆向先求最长下降子序列的长度f[i],就可以知

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5493 题目大意: N个人,每个人有一个唯一的高度h,还有一个排名r,表示它前面或后面比它高的人的个数,求按身高字典序最小同时满足排名的身高排列. 题目思路: [线段树] 首先可以知道,一个人前面或后面有r个人比他高,那么他是第r+1高或第n-i-r+1高,i为这个人是第几高的. 所以先将人按照身高从小到大排序,接下来,把当前这个人放在第k=min(r+1,n-i-r+1)高的位置. 用线段树维护包

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5492 题目大意: 一个N*M的矩阵,一个人从(1,1)走到(N,M),每次只能向下或向右走.求(N+M-1)ΣN+M-1(Ai-Aavg)2最小.Aavg为平均值. (N,M<=30,矩阵里的元素0<=C<=30) 题目思路: [动态规划] 首先化简式子,得原式=(N+M-1)ΣN+M-1(Ai2)-(ΣN+M-1Ai)2 f[i][j][k]表示走到A[i][j]格子上,此时前i+j-1