Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13273    Accepted Submission(s): 6288 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…

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…

vector邻接表: ; struct Edge{ int u,v,w; Edge(int _u=0,int _v=0,int _w=0){u=_u,v=_v,w=_w;} }; vector<Edge> E; vector<int> G[maxn]; void init(int l,int r) { E.clear(); for(int i=l;i<=r;i++) G[i].clear(); } void addedge(int u,int v,int w) { E.pus…

实现功能:同前 程序还是一如既往的优美,虽然比起邻接矩阵的稍稍长了那么些,不过没关系这是必然,但更重要的一个必然是——速度将是一个质的飞跃^_^(这里面的point指针稍作了些创新——anti指针,这个指向当前弧的反向弧,便于路径增广时的操作,相比非递归里面非要用一个op函数来挨个找已经强多了!!!) type point=^node; node=record g,w:longint; anti,next:point; end; var i,j,k,l,m,n,ans,s,t:longint;…

#include<cstdio> #include<algorithm> #include<cstring> #include<queue> #define N 250 #define M 250 #define INF 100000000 using namespace std; int head[N],cur[N],lev[N],ecnt=1,S,T,n,m; queue <int> q; struct adj { int nxt,v,w;…

adj_list_network_edge.h // 邻接表网边数据类模板 template <class WeightType> class AdjListNetworkEdge { public: // 数据成员: int adjVex; // 邻接点 WeightType weight; // 权值 // 构造函数模板: AdjListNetworkEdge(); // 无参数的构造函数模板 AdjListNetworkEdge(int v, WeightType w); // 构造邻接…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14721    Accepted Submission(s): 6968 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…

最大流模板题 大部分Edmond-Karp算法代码都是邻接矩阵实现,试着改成了邻接表. #include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; // 裸最大流 const int N = 2005; const int M = 2005; const int INF = 0x7fffffff; // 邻接表 struct Ed…

最大流裸题,贴下模版 view code#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <queue> using namespace std; const int INF = 0x3f3f3f3f; const int maxn…

解题关键:使用的挑战程序设计竞赛上的模板,第一道网络流题目,效率比较低,且用不习惯的vector来建图. 看到网上其他人说此题有重边,需要注意下,此问题只在邻接矩阵建图时会出问题,邻接表不会存在的,也体现了邻接表的优越性? edge结构体的第三个变量为from的下标. 模板一: #include<bits/stdc++.h> #define MAX_V 17 #define inf 0x3f3f3f3f using namespace std; typedef long long ll; in…