Control
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2139 Accepted Submission(s): 904
Problem Description
You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD
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
Input
There are several test cases.
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).
Output
For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.
See samples for detailed information.
Sample Input
5 6 5 3 5 2 3 4 12 1 5 5 4 2 3 2 4 4 3 2 1
Sample Output
3
Source
2012 ACM/ICPC Asia Regional Chengdu Online
题意：给一个无向图，有些不法分子要从vs点vt点，现在要抓住所有的不法分子阻止他们去vt，那么就要控制某一些城市等待他们，控制每个城市花费不同，问最少花费是多少。
解题：最小割，割断所有的通路，花费使得最少，这样就一定能抓住所有的不法分子。拆点，每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的差别就是不用预处理
*/
#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(或可认为是高度)点的个数
int dis[MAXN]; //每个点到终点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 三个参数，无向边4个参数
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是包括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){ //判断从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; //判断能否从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;
}
//如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）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]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径
//因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流
dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，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);
}
}
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