Tricks Device 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5294 Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu's at the entrance of the tomb while Dumb Zhang's at the end of it. The tomb is made up of many chambers, th…

题意:给一个无向图(连通的),张在第n个点,吴在第1个点,‘吴’只能通过最短路才能到达‘张’,两个问题:(1)张最少毁掉多少条边后,吴不可到达张(2)吴在张毁掉最多多少条边后仍能到达张. 思路:注意是最短路才可达,但是最短路径可能有多条(即权值相等的)!! 第二个问题好回答,来次最短路,记录下到达每个点在最低权值的情况下的最少次用边. 第一个问题,同样只要砍掉最短路的某些边即可.要根据第2个问题所跑的SSSP,将不是最短路的边的剔除,重新建图,跑最大流,得到结果. 当然要考虑重边! #inclu…

题目链接:HDU 5294 Tricks Device 题意:n个点,m条边.而且一个人从1走到n仅仅会走1到n的最短路径.问至少破坏几条边使原图的最短路不存在.最多破坏几条边使原图的最短路劲仍存在 思路: 1.跑一遍最短路.记录下全部最短路径,将这些最短路径的边以(0,1)(流量,容量)加到网络流中,跑一遍最大流 2.记录下的全部最短路径,再加到新的最短路的图中,边权为1,跑一遍最短路.m-最短路 就是答案 注意:有重边的情况 比方: 2 3 1 2 1 1 2 1 1 2 1 ans: 3…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5294 Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 375    Accepted Submission(s): 98 Problem Description Innocent Wu follows Dumb Zh…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5294 Problem Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu's at the entrance of the tomb while Dumb Zhang's at the end of it. The tomb is made up of many chambers, the total…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5294 题意: 给你个无向图: 1.求最少删除几条边就能破坏节点1到节点n的最短路径, 2.最多能删除多少条边同时保证1到n的最短距离不变. 题解: 首先用spfa或dijcstra跑出所有最短路组成的DAG图. 用这个图跑最大流节能解决第一个问题,用这个图跑一遍bfs最短路就能解决第二个问题. 然而我在跑最大流的时候竟然把DAG图建成双向的了orz.. 代码: #include<iostream>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5294   题意:给你n个墓室,m条路径,一个人在1号墓室(起点),另一个人在n号墓室(终点),起点的那个人只有通过最短路径才能追上终点的那个人,而终点的那个人能切断任意路径. 第一问——终点那人要使起点那人不能追上的情况下可以切的最少的路径数,输出最少的路径数 第二问——起点那人能追上终点那人的情况下,终点那人能切断的最多的路径数,输出最多的路径数 思路:要使起点那人无法追上,只要使他的最短路径不存…

题目传送门 /* 题意:一无向图,问至少要割掉几条边破坏最短路,问最多能割掉几条边还能保持最短路 SPFA+Dinic:SPFA求最短路时,用cnt[i]记录到i最少要几条边,第二个答案是m - cnt[n] 最大流==最小割,套个Dinic模板,以后再理解算法... */ #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <vecto…

最短路记录路径,同一时候求出最短的路径上最少要有多少条边, 然后用在最短路上的边又一次构图后求最小割. Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1584    Accepted Submission(s): 388 Problem Description Innocent Wu follows Dum…

第一次做最大流的题目- 这题就是堆模板 #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <cstdio> #include <cstdlib> #include <vector> #include <queue> using namespace std; typedef long long…

Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 124    Accepted Submission(s): 27 Problem Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu's at the entr…

Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 389    Accepted Submission(s): 100 Problem Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu's at the ent…

Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance…

HDU 3605 Escape (网络流,最大流,位运算压缩) Description 2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want you…

HDU 4289 Control (网络流,最大流) Description You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD(Weapon of Mass Destruction)from one city (the source) to another o…

HDU 4292 Food (网络流,最大流) Description You, a part-time dining service worker in your college's dining hall, are now confused with a new problem: serve as many people as possible. The issue comes up as people in your college are more and more difficult…

分析题意可知: 1.最少须要切断多少边使吴不能找到张(题意吴仅仅能走最短路径上面的边),对从起点到终点的最短路径又一次建图,每条边的权值为1.求最大流就可以 2.在吴能够找到张的前提下,最多能够切断边的数量:仅仅须要在全部最短路径中寻找一条经过边数量最少的最短路径,用边的总数减去它就可以(逆向思维) 代码例如以下: #include <cstdio> #include <cstring> #include <vector> #include <queue>…

处理处最短路径图,这个比较巧妙 链接:点我…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3572 Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph. Input The first line of input contains an integer…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5637 题意: http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=675&pid=1003 题解: 令n=(1<<17)-1. 首先很容易想到建图,跑最短路,不过有多次查询,如果每次都跑最短路的话要m*n*logn,会t. 所以可能是我们模型建的太一般化了,需要考虑题目的特殊性. s到t的最少变化次数本质…

转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove 题目:给出n个城市需要去占领,有m条线段是障碍物,有p个士兵可以用.占领城市有个先后顺序,每个士兵有个背包,占领城市之后,仅能补给一次背包.问背包容量最少是多少,可以用这P个士兵完成任务,起点任意 . http://acm.hdu.edu.cn/showproblem.php?pid=4606 首先:枚举所有顶点,求一下距离,判断是否与线段相交.然…

题目地址:HDU 3832 这个题的这种方法我无法给出证明. 我当时这个灵感出来的时候是想的是要想覆盖的点最少,那就要尽量反复利用这些点,然后要有两个之间是通过还有一个点间接连接的,这样会充分利用那些点. 然后就这样写了一次,一直WA..然后中午睡觉的时候突然想到了有一种情况这样做是不正确的.那就是有个点作为中间点,与三个点相连的情况,这样的情况尽管也符合.可是会有反复边... 可是恰恰相反..反复边应该越多越好. . 那就充分利用了这些点.那么这个点应该怎么找呢?那就直接枚举好了.可是枚举每一…

题目描述 有一个\(n\)行\(m\)列的网格图. \(S\)到第一行的每一个点都有一条单向边,容量为\(\infty\). 最后一行的每个点到\(T\)都有一条单向边,容量为\(\infty\). 同一行中相邻的两个节点之间有一条无向边,\((x,y)\)和\((x,y+1)\)之间的无向边的容量为\(a_{x,y}\). 同一列中相邻的两个节点之间有两条有向边,\((x,y)\)到\((x+1,y)\)这条有向边的容量为\(b_{x,y}\),\((x+1,y)\)到\((x,y)\)这条有…

[题目链接](http://acm.hdu.edu.cn/showproblem.ph Problem Description Given a n/n matrix Cij (1<=i,j<=n),We want to find a n/n matrix Xij (1<=i,j<=n),which is 0 or 1. Besides,Xij meets the following conditions: 1.X12+X13+...X1n=1 2.X1n+X2n+...Xn-1n=…

题目链接:[http://acm.hdu.edu.cn/showproblem.php?pid=3339] In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5951    Accepted Submission(s): 1998 Problem Description Since 1945, when the firs…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6071 [题目大意] 给出四个点1,2,3,4,1和2,2和3,3和4,4和1 之间有路相连, 现在从2点出发,最后回到2点,要求路径大于等于K,问路径长度最短是多少 [题解] 取一条与2相连的权值最小的边w. 若存在一条从起点到终点的长度为k的路径, 那么必然存在一条长度为k+2w的路径,只要一开始在那条边上往返走就好了. 设dij表示从起点到i,路径长度模2w为j时,路径长度的最小值. 用最短…

题意:给一个无向图,给起点s,终点t,求最少拆掉几条边使得s到不了t,最多拆几条边使得s能到t 思路: 先跑一边最短路,记录最短路中最短的边数.总边数-最短边数就是第二个答案 第一个答案就是在最短路里面求最小割,也就是求最大流,然后依据最短路在建个新图.权为1.跑一边网络流 模板题.以后就用这套模板了 #include <iostream> #include <cstdio> #include <cstring> #include <queue> #incl…

总算今天静下心来学算法.. Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connect…

http://acm.hdu.edu.cn/showproblem.php?pid=5521 Meeting Problem Description   Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated hisfences they were separated into different blocks. John's farm are divided into …