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…

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…

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…

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): 1879    Accepted Submission(s): 938 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…

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…

奔小康赚大钱 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1836    Accepted Submission(s): 798 Problem Description 传说在遥远的地方有一个非常富裕的村落,有一天,村长决定进行制度改革:重新分配房子.这可是一件大事,关系到人民的住房问题啊.村里共有n间房间,刚好有n家老百姓,考虑到每…

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:poj3565 Ants km算法求最小权完美匹配,浮点权值. 链接:http://poj.org/problem?id=3565 题意:给定n个白点的二维坐标,n个黑点的二维坐标. 求是否存在n条边,每条边恰好连一个白点,一个黑点,且所有的边不相交. 输出所有黑点连接的白点编号. 思路:最小权完美匹配. 假定有白点1(a1,b1), 2(a2,b2), 黑点3(a3,b3),4(a4,b4); 如果1(a1,b1)与3(a3,b3)相连,2(a2,b2)与4(a4,b4)相连,如…

uva 1411 Ants Description Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple trees. Each ant colony needs its own apple tree to feed itself. Bill has a map with coordinates of n ant colonies and n apple tre…

题意: 给出n个城市和m条路,每个城市只能经过一次,想要旅游所有的城市,求需要的最小花费(路径的长度). 分析: 做题之前,首先要知道什么是完美匹配.不然题目做了却不知道为什么可以用这个方法来做.完美匹配{X,Y| E},X.Y集合都有n个点(必须相等),它们必须一对一的匹配,并且所有点都要匹配. 对于此题,每个点都有且只有走一次.把每个点都拆为 i与 i'两个点,i值负责出边(就是i点只有出度),i'负责入边.这样就有了两个集合.集合内的点不会有联系.集合之间的点有联系,但是最后只有是一一对应…

http://acm.hdu.edu.cn/showproblem.php?pid=3488 题意: 给出n个点和m条边,每条边有距离,把这n个点分成1个或多个环,且每个点只能在一个环中,保证有解. 思路: 把一个点分成两部分,1~n和n+i~2*n. 连边的情况是这样的,(src,i,1,0),(i+n,dst,1,0). 如果两个点之间相同,则(i,j+n,1,d). 其实这道题目就是选n条边,如何使得权值之和最小. 具体请参考这http://blog.csdn.net/u013480600…

传送门:http://poj.org/problem?id=3565 Ants Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7650   Accepted: 2424   Special Judge Description Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple tree…

传送门:http://poj.org/problem?id=2195 Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26151   Accepted: 13117 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit st…

题目链接: https://vjudge.net/problem/POJ-2195 题目大意: 给定一个N*M的地图,地图上有若干个man和house,且man与house的数量一致.man每移动一格需花费$1(即单位费用=单位距离),一间house只能入住一个man.现在要求所有的man都入住house,求最小费用. 思路: KM算法传送门: 理解篇    运用篇 每个man和house建立带权二分图,曼哈顿距离就是边的值,这里要求最小费用,也就是二分图最小权值匹配,但是KM算法求的是二分图最…

题意:给定N个点,M条边,M >= N-1.已知M条边都有一个权值,已知前N-1边能构成一颗N个节点生成树,现问通过修改这些边的权值使得最小生成树为前N条边的最小改动总和为多少? 分析:由于计算的最小改动且为最小生成树则显然前N-1条边肯定权值都减少,后面的边权值都增加.由于选择的边为前N-1得到最小生成树,因此首先将N-1条边构图,然后对后面的每一条边,那么这条边所构成的环中,有任意一条边的a与该边b,设原始权重为w[a],w[b],改变量为d[a],w[b],则有w[a] - d[a] <…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3790 Problem Description 给你n个点,m条无向边,每条边都有长度d和花费p,给你起点s终点t,要求输出起点到终点的最短距离及其花费,如果最短距离有多条路线,则输出花费最少的.   Input 输入n,m,点的编号是1~n,然后是m行,每行4个数 a,b,d,p,表示a和b之间有一条边,且其长度为d,花费为p.最后一行是两个数 s,t;起点s,终点.n和m为0时输入结束.(1<n…

题目链接: http://poj.org/problem?id=1797 Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has buil…

本题思路:最短路变形,改变松弛方式即可,dist存的是源结点到当前结点的最长路的最小权值. 参考代码: #include <cstdio> #include <cstring> #include <algorithm> #define INF 0x3f3f3f3f using namespace std; + ; , G[maxn][maxn], dist[maxn]; bool vis[maxn]; int Dijkstra(int source, int aid)…

Sample Input2210 10 //坐标20 2031 12 21000 1000 Sample Output1414.2   //最小权值和*100  保留1位小数oh!       //不连通 prim # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # define LL long…

该题是最小生成树问题变通活用,表示自己开始没有想到该算法:先将所有边按权重排序,然后枚举最小边,求最小生成树(一个简单图的最小生成树的最大权是所有生成树中最大权最小的,这个容易理解,所以每次取最小边,求一次最小生成树,这样差值都次这次最小的),记录更新即可.并查集来判断连通. 类似一提,hoj1598,开始时用DFS搜索,TLE,受启发,用枚举方法差不多,只是在每次枚举最小边的时候结束条件改为起点与终点连通,连通就结束(father(start)==father(end)). #include<…

$ POJ~2018~Best~Cow~ Fences $(二分答案构造新权值) $ solution: $ 题目大意: 给定正整数数列 $ A $ ,求一个平均数最大的长度不小于 $ L $ 的子段 这道题首先我们如果没有长度限制,直接扫一遍数组即可 而有了长度限制之后我们的候选集合发生改变,很容让我们想到DP 事实上这一道题可以DP,用斜率优化复杂度极小,就是有点常数(事实上最优) 但是我们可以参考类似01规划的做法,因为答案具有单调行. 我们让数组中每一个数都减去我们二分答案枚举的值,然后…

http://acm.hdu.edu.cn/showproblem.php?pid=3488 给一个无源汇的,带有边权的有向图 让你找出一个最小的哈密顿回路 可以用KM算法写,但是费用流也行 思路 1. 哈密顿回路对于每个点的流量有限制,因此$V$拆开为$V$和$V'$ 2. 我们建立附加源点$S$和附加汇点$T$哈密顿回路中的每个点有其唯一的后继和前驱,换句话说,对于任意一个点$V$,它满足$in(V)=out(V)$ 为了满足该条件,从源点向$V$ 连接容量为1,费用为0的边,从$V'$向汇…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1281 棋盘游戏 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2905    Accepted Submission(s): 1702 Problem Description 小希和Gardon在玩一个游戏:对一个N*M的棋盘,在格子里放尽…

//是象棋里的车 符合二分匹配 # include<stdio.h> # include<algorithm> # include<string.h> using namespace std; int n,m,pp[110][110],map[110],vis[110]; int bfs(int x) { for(int i=1;i<=m;i++) { if(!vis[i]&&pp[x][i]) { vis[i]=1; if(!map[i]||bf…

Fire Net Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7998    Accepted Submission(s): 4573 Problem Description Suppose that we have a square city with straight streets. A map of a city is a s…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 28359   Accepted: 9213 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)…

给定一棵N个节点的树,编号1~N.其中1号节点是根,并且第i个节点的权值是Vi. 针对这棵树,小Hi会询问小Ho一系列问题.每次小Hi会指定一个节点x,询问小Ho以x为根的子树中,最小的权值是多少.为了增加难度,小Hi可能随时改变其中每个节点的权值. 你能帮助小Ho准确.快速的回答小Hi的问题吗? Input 第一行一个正整数N. 第二行N个整数,V1, V2, ... VN. 第三行n-1个正整数,第i个数Pi表示第i+1号节点的父结点是第Pi号节点.注意1号节点是根. 第四行一个正整数Q,表…

描述 给定一棵N个节点的树,编号1~N.其中1号节点是根,并且第i个节点的权值是Vi. 针对这棵树,小Hi会询问小Ho一系列问题.每次小Hi会指定一个节点x,询问小Ho以x为根的子树中,最小的权值是多少.为了增加难度,小Hi可能随时改变其中每个节点的权值. 你能帮助小Ho准确.快速的回答小Hi的问题吗? 输入 第一行一个正整数N. 第二行N个整数,V1, V2, ... VN. 第三行n-1个正整数,第i个数Pi表示第i+1号节点的父结点是第Pi号节点.注意1号节点是根. 第四行一个正整数Q,表…