Control(拆点+最大流)
Control
http://acm.hdu.edu.cn/showproblem.php?pid=4289
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4995 Accepted Submission(s): 2057
The highway network consists of bidirectional highways, connecting two distinct city. A vehicle can only enter/exit the highway network at cities only.
You may locate some SA (special agents) in some selected cities, so that when the terrorists enter a city under observation (that is, SA is in this city), they would be caught immediately.
It is possible to locate SA in all cities, but since controlling a city with SA may cost your department a certain amount of money, which might vary from city to city, and your budget might not be able to bear the full cost of controlling all cities, you must identify a set of cities, that:
* all traffic of the terrorists must pass at least one city of the set.
* sum of cost of controlling all cities in the set is minimal.
You may assume that it is always possible to get from source of the terrorists to their destination.
------------------------------------------------------------
1 Weapon of Mass Destruction
The first line of a single test case contains two integer N and M ( 2 <= N <= 200; 1 <= M <= 20000), the number of cities and the number of highways. Cities are numbered from 1 to N.
The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.
The following N lines contains costs. Of these lines the ith one contains exactly one integer, the cost of locating SA in the ith city to put it under observation. You may assume that the cost is positive and not exceeding 107.
The followingM lines tells you about highway network. Each of these lines contains two integers A and B, indicating a bidirectional highway between A and B.
Please process until EOF (End Of File).
See samples for detailed information.
5 6
5 3
5
2
3
4
12
1 5
5 4
2 3
2 4
4 3
2 1
拆点+最大流
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define maxn 200005
#define MAXN 200005
#define mem(a,b) memset(a,b,sizeof(a))
const int N=;
const int M=;
const int INF=0x3f3f3f3f;
using namespace std;
int n;
struct Edge{
int v,next;
int cap,flow;
}edge[MAXN*];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
int cnt=;//实际存储总边数
void isap_init()
{
cnt=;
memset(pre,-,sizeof(pre));
}
void isap_add(int u,int v,int w)//加边
{
edge[cnt].v=v;
edge[cnt].cap=w;
edge[cnt].flow=;
edge[cnt].next=pre[u];
pre[u]=cnt++;
}
void add(int u,int v,int w){
isap_add(u,v,w);
isap_add(v,u,);
}
bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里,但是好像是时间要长
{
memset(dep,-,sizeof(dep));
memset(gap,,sizeof(gap));
gap[]=;
dep[t]=;
queue<int>q;
while(!q.empty())
q.pop();
q.push(t);//从汇点开始反向建层次图
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=pre[u];i!=-;i=edge[i].next)
{
int v=edge[i].v;
if(dep[v]==-&&edge[i^].cap>edge[i^].flow)//注意是从汇点反向bfs,但应该判断正向弧的余量
{
dep[v]=dep[u]+;
gap[dep[v]]++;
q.push(v);
//if(v==sp)//感觉这两句优化加了一般没错,但是有的题可能会错,所以还是注释出来,到时候视情况而定
//break;
}
}
}
return dep[s]!=-;
}
int isap(int s,int t)
{
if(!bfs(s,t))
return ;
memcpy(cur,pre,sizeof(pre));
//for(int i=1;i<=n;i++)
//cout<<"cur "<<cur[i]<<endl;
int u=s;
path[u]=-;
int ans=;
while(dep[s]<n)//迭代寻找增广路,n为节点数
{
if(u==t)
{
int f=INF;
for(int i=path[u];i!=-;i=path[edge[i^].v])//修改找到的增广路
f=min(f,edge[i].cap-edge[i].flow);
for(int i=path[u];i!=-;i=path[edge[i^].v])
{
edge[i].flow+=f;
edge[i^].flow-=f;
}
ans+=f;
u=s;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];i!=-;i=edge[i].next)
{
v=edge[i].v;
if(dep[v]+==dep[u]&&edge[i].cap-edge[i].flow)
{
cur[u]=path[v]=i;//当前弧优化
flag=true;
break;
}
}
if(flag)
{
u=v;
continue;
}
int x=n;
if(!(--gap[dep[u]]))return ans;//gap优化
for(int i=pre[u];i!=-;i=edge[i].next)
{
if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
{
x=dep[edge[i].v];
cur[u]=i;//常数优化
}
}
dep[u]=x+;
gap[dep[u]]++;
if(u!=s)//当前点没有增广路则后退一个点
u=edge[path[u]^].v;
}
return ans;
} int main(){
std::ios::sync_with_stdio(false);
int m,s,t;
while(cin>>n>>m){
cin>>s>>t;
t+=n;
int a,b,c;
isap_init();
for(int i=;i<=n;i++){
cin>>c;
add(i,i+n,c);
}
for(int i=;i<=m;i++){
cin>>a>>b;
add(a+n,b,INF);
add(b+n,a,INF);
}
n=n+n;
cout<<isap(s,t)<<endl;
}
}
Control(拆点+最大流)的更多相关文章
- hdu4289 最小割最大流 (拆点最大流)
最小割最大流定理:(参考刘汝佳p369)增广路算法结束时,令已标号结点(a[u]>0的结点)集合为S,其他结点集合为T=V-S,则(S,T)是图的s-t最小割. Problem Descript ...
- BZOJ 1877 晨跑 拆点费用流
题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=1877 题目大意: Elaxia最近迷恋上了空手道,他为自己设定了一套健身计划,比如俯卧 ...
- Risk UVA - 12264 拆点法+最大流+二分 最少流量的节点流量尽量多。
/** 题目:Risk UVA - 12264 链接:https://vjudge.net/problem/UVA-12264 题意:给n个点的无权无向图(n<=100),每个点有一个非负数ai ...
- HDU 3572 Task Schedule(拆点+最大流dinic)
Task Schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) To ...
- POJ 2391 Ombrophobic Bovines ★(Floyd+二分+拆点+最大流)
[题意]有n块草地,一些奶牛在草地上吃草,草地间有m条路,一些草地上有避雨点,每个避雨点能容纳的奶牛是有限的,给出通过每条路的时间,问最少需要多少时间能让所有奶牛进入一个避雨点. 和POJ2112很类 ...
- CF 277E Binary Tree on Plane (拆点 + 费用流) (KM也可做)
题目大意: 平面上有n个点,两两不同.现在给出二叉树的定义,要求树边一定是从上指向下,即从y坐标大的点指向小的点,并且每个结点至多有两个儿子.现在让你求给出的这些点是否能构成一棵二叉树,如果能,使二叉 ...
- 【拆点费用流】【HDU1853】【 Cyclic Tour】
题意: 有N个城市,M条单向路,Tom想环游全部城市,每次至少环游2个城市,每个城市只能被环游一次.由于每条单向路都有长度,要求游遍全部城市的最小长度. // 给定一个有向图,必须用若干个环来覆盖整个 ...
- [SPOJ962]Intergalactic Map 拆点+最大流
Jedi knights, Qui-Gon Jinn and his young apprentice Obi-Wan Kenobi, are entrusted by Queen Padmé Ami ...
- poj-3281(拆点+最大流)
题意:有n头牛,f种食物,d种饮料,每头牛有自己喜欢的食物和饮料,问你最多能够几头牛搭配好,每种食物或者饮料只能一头牛享用: 解题思路:把牛拆点,因为流过牛的流量是由限制的,只能为1,然后,食物和牛的 ...
随机推荐
- HDU 2063 过山车(匈牙利算法)
过山车 Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submissio ...
- Js/jquery获取当前日期时间及其它操作
var myDate = new Date(); myDate.getYear(); //获取当前年份(2位) myDate.getFullYear(); //获取完整的年份(4位,1970-???? ...
- Linux性能分析 vmstat输出
vmstat输出 1.linux系统下vmstat输出 vmstat的输出分为以下几种模式: (1).VM MODE (普通选项) (2).DISK MODE(-d选项) (3).DI ...
- python-pycharm中使用anaconda部署python环境
pycharm中使用anaconda部署python环境 今天来说一下python中一个管理包很好用的工具anaconda,可以轻松实现python中各种包的管理.相信大家都会有这种体验,在pycha ...
- ORM PetaPoco 框架的 CRUD 操作
PetaPoco 的查询操作 public IEnumerable<T> GetAll(string sqlString, object[] obj) { try { IEnumerabl ...
- 并发工具类(四)线程间的交换数据 Exchanger
前言 JDK中为了处理线程之间的同步问题,除了提供锁机制之外,还提供了几个非常有用的并发工具类:CountDownLatch.CyclicBarrier.Semphore.Exchanger.Ph ...
- JPA和Hibernate到底是什么关系???
转自:https://www.cnblogs.com/mosoner/p/9494250.html 在学习框架的过程中,发现学的东西很多,但是感觉他们之间的联系区别都不是很了解,知道JPA可以去实现持 ...
- linux 下常用部分命令
关机 (系统的关机.重启以及登出 ) shutdown -h now 关闭系统() init 关闭系统() shutdown -h hours:minutes & 按预定时间关闭系统 shut ...
- plsql 粘贴
plsql 粘贴
- xe DateTimePicker.Date bug
xe6 bug xe7 ok DateTimePicker1->DateTime.DateString(); DateTimePicker1->DateTime.DateTimeStrin ...