Drainage Ditches

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 65146 Accepted: 25112


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 that Bessie’s clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.

Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.

Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.


The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.


For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 4

1 2 40

1 4 20

2 4 20

2 3 30

3 4 10

Sample Output





using namespace std;
#define maxn 300
int dis[maxn]={0};
int way[maxn][maxn]={0};
int q[maxn*10],h,t;
int n,m,ans;

bool bfs()
    while (h<t)
            int j=q[++h];
            for (int i=1; i<=n; i++)
                    if (dis[i]<0 && way[j][i]>0)
    if (dis[n]>0)
        return true;
        return false;

int dfs(int loc,int low)
    int ans=0;
    if (loc==n) return low;
    for (int i=1; i<=n; i++)
        if (way[loc][i]>0 && dis[i]==dis[loc]+1 && (ans=dfs(i,min(loc,way[loc][i]))))
                return ans;
    return 0;

int main()
    while (~scanf("%d%d",&m,&n))
            for (int i=1; i<=m; i++)
                    int s,e,water;
            while (bfs())
                    int now;
                    while (now=dfs(1,0x7fffffff))
    return 0;

POJ-1273 Drainage Ditches 最大流Dinic的更多相关文章

  1. POJ 1273 Drainage Ditches(最大流Dinic 模板)

    #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, ...

  2. POJ 1273 - Drainage Ditches - [最大流模板题] - [EK算法模板][Dinic算法模板 - 邻接表型]

    题目链接:http://poj.org/problem?id=1273 Time Limit: 1000MS Memory Limit: 10000K Description Every time i ...

  3. poj 1273 Drainage Ditches 最大流入门题

    题目链接:http://poj.org/problem?id=1273 Every time it rains on Farmer John's fields, a pond forms over B ...

  4. POJ 1273 Drainage Ditches(网络流dinic算法模板)

    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 <stdi ...

  5. Poj 1273 Drainage Ditches(最大流 Edmonds-Karp )

    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 ...

  6. POJ 1273 Drainage Ditches 最大流

    这道题用dinic会超时 用E_K就没问题 注意输入数据有重边.POJ1273 dinic的复杂度为O(N*N*M)E_K的复杂度为O(N*M*M)对于这道题,复杂度是相同的. 然而dinic主要依靠 ...

  7. POJ 1273 Drainage Ditches | 最大流模板

    #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #defi ...

  8. poj 1273 Drainage Ditches(最大流)

    http://poj.org/problem?id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Subm ...

  9. POJ 1273 Drainage Ditches (网络最大流)

    http://poj.org/problem? id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Sub ...

  10. POJ 1273 Drainage Ditches(网络流,最大流)

    Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover ...


  1. MongoVue中Collections无法显示的问题

    问题描述: 通过Python向MongoDB写入数据后,MongoVue中Collections无法显示的问题 原因: Mongodb 3.0之后默认的 storageEngine为wiredTige ...

  2. BZOJ2818 Gcd

    本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000作者博客:http://www.cnblogs.com/ljh2000-jump/转 ...

  3. python 国内源

    pipy国内镜像目前有: http://pypi.douban.com/  豆瓣 http://pypi.hustunique.com/  华中理工大学 http://pypi.sdutlinux.o ...

  4. Mina工作原理分析

    Mina是Apache社区维护的一个开源的高性能IO框架,在业界内久经考验,广为使用.Mina与后来兴起的高性能IO新贵Netty一样,都是韩国人Trustin Lee的大作,二者的设计理念是极为相似 ...

  5. SqlParameter设定value为0却变成null

    直接MSDN:http://msdn.microsoft.com/zh-cn/library/0881fz2y(VS.80).aspx 当在 value 参数中指定 Object 时,SqlDbTyp ...

  6. OpenGL学习之路(四)

    1 引子 上次读书笔记主要是学习了应用三维坐标变换矩阵对二维的图形进行变换,并附带介绍了GLSL语言的编译.链接相关的知识,之后介绍了GLSL中变量的修饰符,着重介绍了uniform修饰符,来向着色器 ...

  7. Orcle数据库查询练习复习:四

    一.题目 1.找出张三的最高分和最低分以及对应的课程名 select * from course c,mark m where c.cid=m.cid and sid =(select sid fro ...


    SQL UNION 和 UNION ALL 操作符 SQL Full Join SQL Select Into SQL UNION 操作符 UNION 操作符用于合并两个或多个 SELECT 语句的结 ...

  9. git extrad_addons 部署说明

    注册一个git账号 : 网址:  https://github.com/ 1:安装git   sudo apt-get install git 2:  b把urc扩展占模块pull下来    cd   ...

  10. IC卡接口芯片TDA8007的读写器设计

    摘要:阐述T=0传输协议,给出IC卡读写器中使用的IC卡APDU指令流程和原理框图:重点介绍其中的IC卡接口芯片Philips的TDA8007,给出通过TDA8007对CPU IC卡上下电过程.具体程 ...