Control(拆點+最大流)
阿新 • • 發佈:2018-11-12
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.
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1
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
Source 2012 ACM/ICPC Asia Regional Chengdu Online
拆點+最大流
1 #include<iostream> 2 #include<cstring> 3 #include<string> 4 #include<cmath> 5 #include<cstdio> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #include<set> 10 #define maxn 200005 11 #define MAXN 200005 12 #define mem(a,b) memset(a,b,sizeof(a)) 13 const int N=200005; 14 const int M=200005; 15 const int INF=0x3f3f3f3f; 16 using namespace std; 17 int n; 18 struct Edge{ 19 int v,next; 20 int cap,flow; 21 }edge[MAXN*20];//注意這裡要開的夠大。。不然WA在這裡真的想罵人。。問題是還不報RE。。 22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN]; 23 int cnt=0;//實際儲存總邊數 24 void isap_init() 25 { 26 cnt=0; 27 memset(pre,-1,sizeof(pre)); 28 } 29 void isap_add(int u,int v,int w)//加邊 30 { 31 edge[cnt].v=v; 32 edge[cnt].cap=w; 33 edge[cnt].flow=0; 34 edge[cnt].next=pre[u]; 35 pre[u]=cnt++; 36 } 37 void add(int u,int v,int w){ 38 isap_add(u,v,w); 39 isap_add(v,u,0); 40 } 41 bool bfs(int s,int t)//其實這個bfs可以融合到下面的迭代裡,但是好像是時間要長 42 { 43 memset(dep,-1,sizeof(dep)); 44 memset(gap,0,sizeof(gap)); 45 gap[0]=1; 46 dep[t]=0; 47 queue<int>q; 48 while(!q.empty()) 49 q.pop(); 50 q.push(t);//從匯點開始反向建層次圖 51 while(!q.empty()) 52 { 53 int u=q.front(); 54 q.pop(); 55 for(int i=pre[u];i!=-1;i=edge[i].next) 56 { 57 int v=edge[i].v; 58 if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是從匯點反向bfs,但應該判斷正向弧的餘量 59 { 60 dep[v]=dep[u]+1; 61 gap[dep[v]]++; 62 q.push(v); 63 //if(v==sp)//感覺這兩句優化加了一般沒錯,但是有的題可能會錯,所以還是註釋出來,到時候視情況而定 64 //break; 65 } 66 } 67 } 68 return dep[s]!=-1; 69 } 70 int isap(int s,int t) 71 { 72 if(!bfs(s,t)) 73 return 0; 74 memcpy(cur,pre,sizeof(pre)); 75 //for(int i=1;i<=n;i++) 76 //cout<<"cur "<<cur[i]<<endl; 77 int u=s; 78 path[u]=-1; 79 int ans=0; 80 while(dep[s]<n)//迭代尋找增廣路,n為節點數 81 { 82 if(u==t) 83 { 84 int f=INF; 85 for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增廣路 86 f=min(f,edge[i].cap-edge[i].flow); 87 for(int i=path[u];i!=-1;i=path[edge[i^1].v]) 88 { 89 edge[i].flow+=f; 90 edge[i^1].flow-=f; 91 } 92 ans+=f; 93 u=s; 94 continue; 95 } 96 bool flag=false; 97 int v; 98 for(int i=cur[u];i!=-1;i=edge[i].next) 99 { 100 v=edge[i].v; 101 if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow) 102 { 103 cur[u]=path[v]=i;//當前弧優化 104 flag=true; 105 break; 106 } 107 } 108 if(flag) 109 { 110 u=v; 111 continue; 112 } 113 int x=n; 114 if(!(--gap[dep[u]]))return ans;//gap優化 115 for(int i=pre[u];i!=-1;i=edge[i].next) 116 { 117 if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x) 118 { 119 x=dep[edge[i].v]; 120 cur[u]=i;//常數優化 121 } 122 } 123 dep[u]=x+1; 124 gap[dep[u]]++; 125 if(u!=s)//當前點沒有增廣路則後退一個點 126 u=edge[path[u]^1].v; 127 } 128 return ans; 129 } 130 131 132 int main(){ 133 std::ios::sync_with_stdio(false); 134 int m,s,t; 135 while(cin>>n>>m){ 136 cin>>s>>t; 137 t+=n; 138 int a,b,c; 139 isap_init(); 140 for(int i=1;i<=n;i++){ 141 cin>>c; 142 add(i,i+n,c); 143 } 144 for(int i=1;i<=m;i++){ 145 cin>>a>>b; 146 add(a+n,b,INF); 147 add(b+n,a,INF); 148 } 149 n=n+n; 150 cout<<isap(s,t)<<endl; 151 } 152 }View Code