HDU 4289 Control (最小割 拆点)
Control
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2139 Accepted Submission(s): 904
1 from one city (the source) to another one (the destination). You know their date, source and destination, and they are using the highway network.
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
3
拆点,每一个点拆成一条有向边v->v ’ 边权为控制这个城市的花费,原图中的边u->v,则建成:u+n->v。v+n->u,边权都为INF。再跑一下最大流,就是ans。
/*
最大流:SAP算法,与ISAP的区别就是不用预处理
*/
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define captype int const int MAXN = 100010; //点的总数
const int MAXM = 400010; //边的总数
const int INF = 1<<30;
struct EDG{
int to,next;
captype cap,flow;
} edg[MAXM];
int eid,head[MAXN];
int gap[MAXN]; //每种距离(或可觉得是高度)点的个数
int dis[MAXN]; //每一个点到终点eNode 的最短距离
int cur[MAXN]; //cur[u] 表示从u点出发可流经 cur[u] 号边
int pre[MAXN]; void init(){
eid=0;
memset(head,-1,sizeof(head));
}
//有向边 三个參数。无向边4个參数
void addEdg(int u,int v,captype c,captype rc=0){
edg[eid].to=v; edg[eid].next=head[u];
edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++; edg[eid].to=u; edg[eid].next=head[v];
edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
captype maxFlow_sap(int sNode,int eNode, int n){//n是包含源点和汇点的总点个数。这个一定要注意
memset(gap,0,sizeof(gap));
memset(dis,0,sizeof(dis));
memcpy(cur,head,sizeof(head));
pre[sNode] = -1;
gap[0]=n;
captype ans=0; //最大流
int u=sNode;
while(dis[sNode]<n){ //推断从sNode点有没有流向下一个相邻的点
if(u==eNode){ //找到一条可增流的路
captype Min=INF ;
int inser;
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]) //从这条可增流的路找到最多可增的流量Min
if(Min>edg[i].cap-edg[i].flow){
Min=edg[i].cap-edg[i].flow;
inser=i;
}
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){
edg[i].flow+=Min;
edg[i^1].flow-=Min; //可回流的边的流量
}
ans+=Min;
u=edg[inser^1].to;
continue;
}
bool flag = false; //推断是否能从u点出发可往相邻点流
int v;
for(int i=cur[u]; i!=-1; i=edg[i].next){
v=edg[i].to;
if(edg[i].cap-edg[i].flow>0 && dis[u]==dis[v]+1){
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag){
u=v;
continue;
}
//假设上面没有找到一个可流的相邻点,则改变出发点u的距离(也可觉得是高度)为相邻可流点的最小距离+1
int Mind= n;
for(int i=head[u]; i!=-1; i=edg[i].next)
if(edg[i].cap-edg[i].flow>0 && Mind>dis[edg[i].to]){
Mind=dis[edg[i].to];
cur[u]=i;
}
gap[dis[u]]--;
if(gap[dis[u]]==0) return ans; //当dis[u]这样的距离的点没有了,也就不可能从源点出发找到一条增广流路径
//由于汇点到当前点的距离仅仅有一种。那么从源点到汇点必定经过当前点,然而当前点又没能找到可流向的点,那么必定断流
dis[u]=Mind+1;//假设找到一个可流的相邻点,则距离为相邻点距离+1。假设找不到。则为n+1
gap[dis[u]]++;
if(u!=sNode) u=edg[pre[u]^1].to; //退一条边
}
return ans;
}
int main()
{
int n,m,vs,vt,u,v,cost,ans;
while(scanf("%d%d",&n,&m)>0)
{
scanf("%d%d",&vs,&vt);
vt+=n;
init();
for(int i=1; i<=n; i++){
scanf("%d",&cost);
addEdg(i , i+n , cost);
}
while(m--){
scanf("%d%d",&u,&v);
addEdg(u+n , v , INF);
addEdg(v+n , u , INF);
}
ans=maxFlow_sap(vs , vt , n*2);
printf("%d\n",ans);
}
}
HDU 4289 Control (最小割 拆点)的更多相关文章
- HDU 4289 Control 最小割
Control 题意:有一个犯罪集团要贩卖大规模杀伤武器,从s城运输到t城,现在你是一个特殊部门的长官,可以在城市中布置眼线,但是布施眼线需要花钱,现在问至少要花费多少能使得你及时阻止他们的运输. 题 ...
- hdu-4289.control(最小割 + 拆点)
Control Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Sub ...
- HDU 4289 Control(最大流+拆点,最小割点)
题意: 有一群恐怖分子要从起点st到en城市集合,你要在路程中的城市阻止他们,使得他们全部都被抓到(当然st城市,en城市也可以抓捕).在每一个城市抓捕都有一个花费,你要找到花费最少是多少. 题解: ...
- HDU 4289 Control (网络流,最大流)
HDU 4289 Control (网络流,最大流) Description You, the head of Department of Security, recently received a ...
- hdu 4289 Control(最小割 + 拆点)
http://acm.hdu.edu.cn/showproblem.php?pid=4289 Control Time Limit: 2000/1000 MS (Java/Others) Mem ...
- HDU4289 Control —— 最小割、最大流 、拆点
题目链接:https://vjudge.net/problem/HDU-4289 Control Time Limit: 2000/1000 MS (Java/Others) Memory Li ...
- hdu4289 Control --- 最小割,拆点
给一个无向图.告知敌人的起点和终点.你要在图上某些点安排士兵.使得敌人不管从哪条路走都必须经过士兵. 每一个点安排士兵的花费不同,求最小花费. 分析: 题意可抽象为,求一些点,使得去掉这些点之后,图分 ...
- HDU(2485),最小割最大流
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2485 Destroying the bus stations Time Limit: 40 ...
- HDU 4971 (最小割)
Problem A simple brute force problem (HDU 4971) 题目大意 有n个项目和m个问题,完成每个项目有对应收入,解决每个问题需要对应花费,给出每个项目需解决的问 ...
随机推荐
- git web 服务器的搭建【转】
转自:http://blog.csdn.net/transformer_han/article/details/6450200 目录(?)[-] git服务器搭建过程 需求 硬件需求一台Ubuntu或 ...
- postman接口间关联
现有A.B两个接口,A接口返回的response的body中的部分数据,是B接口发起请求部分数据的来源. A接口response的body结构如下: { "address": &q ...
- python memcache操作-安装、连接memcache
安装memecache wget http://memcached.org/latest tar -zxvf memcached-1.x.x.tar.gz cd memcached-1.x.x ./c ...
- python的算法:二分法查找(1)
1.什么是二分法查找: 1.从数组的中间元素开始,如果中间元素正好是要查找的元素,则搜素过程结束: 2.如果某一特定元素大于或者小于中间元素,则在数组大于或小于中间元素的那一半中查找,而且跟开始一样从 ...
- Android 使用WebView控件展示SVG图
1.添加布局界面代码: <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" xm ...
- 上传文件提示413 Request Entity Too Large错误
打开nginx主配置文件nginx.conf 一般在/usr/local/nginx/conf/nginx.conf这个位置 找到http{}段并修改以下内容 client_max_body_size ...
- Mysql错误:Every derived table must have its own alias
Mysql报错:Every derived table must have its own alias <缺少一个别名> 在子查询中经常会犯的错误 .这个别名其实没啥用途.... ...
- Jmeter 监控远程服务器
压力测试中如何监控服务器的性能呢? 性能选取哪些指标? 环境配置: Jmeter安装路径:E:\SOFEWARE\apache-jmeter-3.1: 需要将JMeterPlugins-Extras. ...
- 【xunsearch】笔记
1.添加索引 $ cd /usr/local/xunsearch/sdk/php/ $ util/Indexer.php --rebuild --source=mysql://数据库用户名:数据库密码 ...
- Tarjan+topsort(DP)【P3387】 [模板]缩点
Description 给定一个n个点m条边有向图,每个点有一个权值,求一条路径,使路径经过的点权值之和最大.你只需要求出这个权值和. 允许多次经过一条边或者一个点,但是,重复经过的点,权值只计算一次 ...