HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some route…

Island Transport Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships. You have a transportation company there. Some routes are opened for p…

Island Transport Time Limit: 10000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 4280 64-bit integer IO format: %I64d      Java class name: Main In the vast waters far far away, there are many islands. People are living on…

HDU 4280:http://acm.hdu.edu.cn/showproblem.php?pid=4280 题意: 比较裸的最大流题目,就是这是个无向图,并且比较卡时间. 思路: 是这样的,由于是无向图,所以addedge 的反边容量直接设为原始流量.然后还可以优化搜索的方向,bfs可以从t到s跑,dfs可以从s到t跑,这样快. //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000&qu…

Island Transport Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 6217    Accepted Submission(s): 1965 Problem Description In the vast waters far far away, there are many islands. People are l…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:pid=4280">http://acm.hdu.edu.cn/showproblem.php?pid=4280 Problem Description In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the is…

题意: 求最大流 思路: \(1e5\)条边,偷了一个超长的\(HLPP\)板子.复杂度\(n^2 \sqrt{m}\).但通常在随机情况下并没有isap快. 板子: template<class T = int> struct HLPP{ const int MAXN = 1e5 + 5; const T INF = 0x3f3f3f3f; struct edge{ int to, rev; T f; }; vector<edge> adj[maxn]; deque<int…

#include<bits/stdc++.h>using namespace std;typedef long long ll;const ll inf=0x3f3f3f3f;int cnt=1;//边的数量int head[110],cur[110];//head记录每一个点最后一条边的编号,cur记录当前点循环到了哪一条边int n,deep[110],s,t,start;//s设为不选,t为选,即源点和汇点//权值为负的点向s连容量为-w的边,权值为正的点向t连容量为w的边//这里的容量…

Problem Description 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. For the i-th task, the factory…

在阅读本文前,建议先自学最大流的Ek算法. 引入 Ek的核心是执行bfs,一旦找到增广路就停下来进行增广.换言之,执行一遍BFS执行一遍DFS,这使得效率大大降低.于是我们可以考虑优化. 核心思路 在一次BFS中,找到的增广路可能不止一条,这时我们可以本着“尽量少进行BFS”的想法,在一次bfs后把所有能增广的路径全部增广.具体怎么做呢?仍然是:while(bfs(源点,汇点)) dfs(): 每次bfs标记出每个点的“深度”,也就是距离源点的长度.我们将得到的新图称作分层图.接下来我们在分层图…