HDU 4289 Control (最小割最大流)

it2024-07-10  6

Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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 10 7.   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   分析 这种就是最小割问题,使源点和汇点不连通。有向图的最小割就是最大流,所以只需要建立有向图跑最大流就行了。 因为给的是城市的代价,所以需要将城市拆分为2个,边权就是代价。另外,已存在的有向边,要从未拆分点连到拆分点(为了经过城市),并将容量设为无限大 代码 #include<stdio.h> #include<string.h> #include<iostream> using namespace std; //**************************************************** //最大流模板 SAP算法 //邻接表形式 //****************************************************** const int MAXN = 400+100;//点数的最大值 const int MAXM = 100000;//边数的最大值 const int INF = 0x3f3f3f3f; struct Node { int from,to,next; int cap; }edge[MAXM]; int tol; int head[MAXN]; int dep[MAXN]; int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y int n;//n是总的点的个数,包括源点和汇点 void init() { tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w) { edge[tol].from=u; edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; head[u]=tol++; edge[tol].from=v; edge[tol].to=u; edge[tol].cap=0; edge[tol].next=head[v]; head[v]=tol++; } void BFS(int start,int end) { memset(dep,-1,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0]=1; int que[MAXN]; int front,rear; front=rear=0; dep[end]=0; que[rear++]=end; while(front!=rear) { int u=que[front++]; if(front==MAXN)front=0; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].to; if(dep[v]!=-1)continue; que[rear++]=v; if(rear==MAXN)rear=0; dep[v]=dep[u]+1; ++gap[dep[v]]; } } } int SAP(int start,int end) { int res=0; BFS(start,end); int cur[MAXN]; int S[MAXN]; int top=0; memcpy(cur,head,sizeof(head)); int u=start; int i; while(dep[start]<n) { if(u==end) { int temp=INF; int inser; for(i=0;i<top;i++) if(temp>edge[S[i]].cap) { temp=edge[S[i]].cap; inser=i; } for(i=0;i<top;i++) { edge[S[i]].cap-=temp; edge[S[i]^1].cap+=temp; } res+=temp; top=inser; u=edge[S[top]].from; } if(u!=end&&gap[dep[u]-1]==0)//出现断层,无增广路 break; for(i=cur[u];i!=-1;i=edge[i].next) if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1) break; if(i!=-1) { cur[u]=i; S[top++]=i; u=edge[i].to; } else { int min=n; for(i=head[u];i!=-1;i=edge[i].next) { if(edge[i].cap==0)continue; if(min>dep[edge[i].to]) { min=dep[edge[i].to]; cur[u]=i; } } --gap[dep[u]]; dep[u]=min+1; ++gap[dep[u]]; if(u!=start)u=edge[S[--top]].from; } } return res; } int main() { int N,M; int u,v; while(scanf("%d%d",&N,&M)!=EOF){ int st,ed; init(); scanf("%d%d",&st,&ed); st=2*st-1; ed=2*ed; n=2*N; for(int i=1;i<=N;i++){ scanf("%d",&u); addedge(2*i-1,2*i,u); addedge(2*i,2*i-1,u); } while(M--){ scanf("%d%d",&u,&v); addedge(2*u,2*v-1,INF); addedge(2*v,2*u-1,INF); } printf("%d\n",SAP(st,ed)); } return 0; }

 

转载于:https://www.cnblogs.com/wangdongkai/p/5611442.html

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