Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so t…

找了好久这两个的区别...UVA820 WA了 好多次.不过以后就做模板了,可以求任意两点之间的最大流. UVA 是无向图,因此可能有重边,POJ 1273是有向图,而且是单源点求最大流,因此改模板的时候注意一下. 而且我居然犯了更愚蠢的错误,以为重边的时候需要选最大的,正解应该是累加.... #include<stdio.h> #include<queue> #include<string.h> #define INF 999999 using namespace s…

POJ 1273给出M条边,N个点,求源点1到汇点N的最大流量. 本文主要就是附上dinic的模板,供以后参考. #include <iostream> #include <stdio.h> #include <algorithm> #include <queue> #include <string.h> /* POJ 1273 dinic算法模板 边是有向的,而且存在重边,且这里重边不是取MAX,而是累加和 */ using namespace…

题目链接:http://poj.org/problem?id=1273 a.EK算法:(Edmond-Karp): 用BFS不断找增广路径,当找不到增广路径时当前流量即为最大流. b.dinic算法:不断找最短路. 题意:现在有m个池塘(从1到m开始编号,1为源点,m为汇点),及n条水渠,给出这n条水渠所连接的池塘和所能流过的水量,求水渠中所能流过的水的最大容量. //http://blog.csdn.net/huzhengnan/article/details/7766446一个写的特别好的博…

题目链接:http://poj.org/problem?id=1273 Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has…

http://poj.org/problem?id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 62708   Accepted: 24150 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means…

题目链接:http://poj.org/problem?id=1273 Time Limit: 1000MS Memory Limit: 10000K Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes…

http://poj.org/problem?id=1273 这道题很值得反思,弄了一下午,交上去先是一直编译错误,而在本地运行没有问题, 原因可能是oj的编译器版本老旧不支持这样的写法 G[from].push_back((edge){to,cap,G[to].size()}); G[to].push_back((edge){from,0,G[from].size() - 1}); 这两句交上去是不能通过的,不知道网上很多人这样子贴出代码是怎样通过的(‘白书’的模板有问题的) 如果是照着‘白书…

链接: http://poj.org/problem?id=1273 代码: //Dinic #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; ; ; struct Edge { int v, flow, next; }edge[MAXN]; int Head[MAXN], cnt; int Layer[MAXN]; vo…

链接: http://poj.org/problem?id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63763   Accepted: 24613 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This m…

http://poj.org/problem? id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 55235   Accepted: 21104 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This mean…

转载 百度文库花了5分下的 不过确实是自己需要的东西经典的最大流题POJ1273 ——其他练习题 POJ3436 . 题意描述: 现在有m个池塘(从1到m开始编号,1为源点,m为汇点),及n条水渠,给出这n条水渠所连接的池塘和所能流过的水量,求水渠中所能流过的水的最大容量.一道基础的最大流题目.但是模板小心使用,目前只是求单源点的最大流. 参考数据: 输入: 5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10 输出: 50 程序实现: 增广路算法Edmonds_Karp…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 69355   Accepted: 26873 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by…

题目链接:poj1273 Drainage Ditches 呜呜,今天自学网络流,看了EK算法,学的晕晕的,留个简单模板题来作纪念... #include<cstdio> #include<vector> #include<queue> #include<cstring> #include<set> #include<algorithm> #define CLR(a,b) memset((a),(b),sizeof((a))) usi…

Drainage Ditches 题目抽象:给你m条边u,v,c.   n个定点,源点1,汇点n.求最大流.  最好的入门题,各种算法都可以拿来练习 (1):  一般增广路算法  ford() #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> #include…

题目大概意思是,有N条水沟和M个水池,问从第一个水池到最后一个水池在同一时间内能够流过多少水第一行有两个整数N,M接下来N行,每行有3个整数,a,b,c,代表从a到b能够流c单位的水超级模板题,一个有向图,源点为1,汇点为M,套最大流模板就行了 我就通过这水题理解EK的... #include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<climits>…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 53640   Accepted: 20448 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by…

题意:给出起点是一个池塘,M条沟渠,给出这M条沟渠的最大流量,再给出终点是一条河流,问从起点通过沟渠最多能够排多少水到河流里面去 看的紫书的最大流,还不是很理解,照着敲了一遍 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #includ…

前言 在开发高并发系统时有三把利器用来保护系统:缓存.降级和限流 缓存: 缓存的目的是提升系统访问速度和增大系统处理容量 降级: 降级是当服务出现问题或者影响到核心流程时,需要暂时屏蔽掉,待高峰或者问题解决后再打开 限流: 限流的目的是通过对并发访问/请求进行限速,或者对一个时间窗口内的请求进行限速来保护系统,一旦达到限制速率则可以拒绝服务.排队或等待.降级等处理 常见限流算法 漏桶算法 漏桶算法思路很简单,水(请求)先进入到漏桶里,漏桶以一定的速度出水,当水流入速度过大会直接溢出,可以看出漏桶…

一 基本知识了解(  字节流, 字符流, byte,bit是啥?) /* java中字节流和字符流之前有接触过,但是一直没有深入的学习和了解. 今天带着几个问题,简单的使用字节流的基本操作. 1 什么是字节流? 字节流是由字节组成的,字符流是由字符组成的.Java里字符由两个字节组成.字节流是最基本的 所有的InputStream和OutputStream的子类都是,主要用在处理二进制数据,它是按字节来处理的但实际中很多的数据是文本, 又提出了字符流的概念,它是按虚拟机的encode来处理, 也…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 67823   Accepted: 26209 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 67387   Accepted: 26035 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by…

一直不明白为什么我的耗时几百毫秒,明明差不多的程序啊,我改来改去还是几百毫秒....一个小时后:明白了,原来把最大值0x3f(77)取0x3f3f3f3f就把时间缩短为16ms了.可是为什么原来那样没有WA呢?哦,明白了,因为最大值写小了,dicnic会多跑几遍.嗯顺便:0x3f3f3f3f也就是1061109567,也就是10^9级别的,是一个很大的数,而且加上一个10^9级别以下的数字也不会溢出.所以很多时候最大值取这个会比较好.最大的int是0x7fffffff,可以用常量INT_MAX…

dinic版本 感觉dinic算法好帅,比Edmonds-Karp算法不知高到哪里去了 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Th…

#include<cstdio> #include<iostream> #include<cstring> #include<queue> #define INF 9999999 #define M 330 using namespace std; int maxflow,pre[M],map[M][M],n,m; void Flow(int start,int end) { ) { queue<int> p; int minflow=INF;…

明天再拍一遍 #include <iostream> #include <queue> using namespace std; ; const int INF = 0x7FFFFFFF; int n,m,map[N][N],path[N],flow[N],start,end; queue<int> q; int bfs() { int i,t; while(!q.empty()) q.pop(); memset(path,-,sizeof(path)); path[s…

点击打开链接 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 49648   Accepted: 18829 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is cove…

虽然算法还没有理解透,但以及迫不及待地想要A道题了. 非常裸的最大流,试试lrj的模板练练手. #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; + ; ; struct Edge { int from, to, cap, flow; Edge(int u, int v,…

Drainage DitchesHal Burch Time Limit 1000 ms Memory Limit 65536 kb description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a…

Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63924   Accepted: 24673 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by…