[Training Questions] exam confidence test is the smallest cut, and training exam

[Training Questions] exam confidence test is the smallest cut, and training exam

Question description:

There are N people in the test room, A and B. If student I is in Test Room A, the total confidence value increases by xi; If student I is in Test Room B, the total confidence value increases by yi. There are also m friends. When I and I are both in the test room, the total confidence value increases ai; if both are in the B test room, the total confidence value increases bi; if two people are in different test rooms, the total confidence is reduced by ci.

Ask the maximum value of the total confidence value (the initial value of the total confidence value is 0 ).

N <= 10000, M <= 50000, time limit = 1 s

 

Analysis:

It is easy to find that this is a problem with network streams, but modeling is a little difficult for people who are not very skilled. For example, at the time of the examination, I felt like a minimal cut, but I thought about how to create a maximum fee, but how to create a graph is cool... ORZ

The maximum fee is a positive idea. If you cannot do it, analyze it. It can be found that the upper limit of the total confidence value is determined at the beginning. When a student determines that he or she is in a certain test room, he or she will lose the benefit of another option, that is, the cost is paid. The minimum cost is equivalent to the maximum benefit, so it can be converted to the minimum cut. Considering that there are only two statuses for each person, in test room A or test room B, the Source and Sink points are set up to represent test room A and test room B respectively. Then we start to have a happy EDGE connection. The significance of an edge capacity is how much it will cost when the edge is cut off. As long as you really understand the idea of using the minimum cut, this question is actually not difficult. (Because of the problem of dividing by 2, the capacity becomes a floating point, so when the edge is connected, all the costs are multiplied by 2)

Since it is really troublesome to explain, I will not repeat the code below.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cstdlib>  5 #include<algorithm>  6 #include<cmath>  7 #include<queue>  8 #include<set>  9 #include<map> 10 #include<vector> 11 #include<cctype> 12 #define inf 9e18  13 using namespace std; 14 const int maxn=10005; 15 const int maxm=50005; 16 typedef long long LL; 17  18 int N,M,A[maxn],B[maxn],S,T,tot; 19 struct data{ int u,v,a,b,c; }C[maxm]; 20 struct net_edge{ int from,to,next; LL cap,flow; }NE[4*maxn+10*maxm]; 21 int nfirst[maxn],nnp,cur[maxn],fl[maxn],d[maxn],gap[maxn]; 22 int mq[maxn],front,rear; 23  24 void _scanf(int &x) 25 { 26     x=0; 27     char ch=getchar(); 28     while(ch<'0'||ch>'9') ch=getchar(); 29     while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); 30 } 31 void data_in() 32 { 33     _scanf(N);_scanf(M); 34     for(int i=1;i<=N;i++) _scanf(A[i]); 35     for(int i=1;i<=N;i++) _scanf(B[i]); 36     for(int i=1;i<=M;i++){ 37         _scanf(C[i].u);_scanf(C[i].v); 38         _scanf(C[i].a);_scanf(C[i].b);_scanf(C[i].c); 39     } 40 } 41 void net_add_edge(int u,int v,LL cap) 42 { 43     NE[++nnp]=(net_edge){u,v,nfirst[u],cap,0}; 44     nfirst[u]=nnp; 45     NE[++nnp]=(net_edge){v,u,nfirst[v],0,0}; 46     nfirst[v]=nnp; 47 } 48 void _net_add_edge(int u,int v,LL cap) 49 { 50     NE[++nnp]=(net_edge){u,v,nfirst[u],cap,0}; 51     nfirst[u]=nnp; 52     NE[++nnp]=(net_edge){v,u,nfirst[v],cap,0}; 53     nfirst[v]=nnp; 54 } 55 void build_net() 56 { 57     S=N+1,T=N+2,tot=T; 58     for(int i=1;i<=N;i++){ 59         net_add_edge(S,i,(LL)2*B[i]); 60         net_add_edge(i,T,(LL)2*A[i]); 61     } 62     int u,v; 63     for(int i=1;i<=M;i++){ 64         u=C[i].u,v=C[i].v; 65         net_add_edge(S,u,(LL)C[i].b+C[i].c); 66         net_add_edge(u,T,(LL)C[i].a+C[i].c); 67         net_add_edge(S,v,(LL)C[i].b+C[i].c); 68         net_add_edge(v,T,(LL)C[i].a+C[i].c); 69         _net_add_edge(u,v,(LL)C[i].a+C[i].b+(LL)2*C[i].c); 70     } 71 } 72 void BFS(int s) 73 { 74     for(int i=1;i<=tot;i++) d[i]=tot; 75     front=rear=0; 76     mq[rear++]=s; 77     d[s]=0; 78     int i,j; 79     while(front!=rear){ 80         i=mq[front++]; 81         for(int p=nfirst[i];p;p=NE[p].next){ 82             j=NE[p].to; 83             if(d[j]==tot) d[j]=d[i]+1,mq[rear++]=j; 84         } 85     } 86 } 87 LL augment(int s,int t) 88 { 89     int now=t; LL flow=inf; 90     while(now!=s){ 91         flow=min(flow,NE[fl[now]].cap-NE[fl[now]].flow); 92         now=NE[fl[now]].from; 93     } 94     now=t; 95     while(now!=s){ 96         NE[fl[now]].flow+=flow,NE[(fl[now]-1^1)+1].flow-=flow; 97         now=NE[fl[now]].from; 98     } 99     return flow;100 }101 LL ISAP(int s,int t)102 {103     memcpy(cur,nfirst,sizeof(cur));104     BFS(t);105     for(int i=1;i<=tot;i++) gap[d[i]]++;106     LL maxflow=0; int now=s,j;107     while(d[s]<tot){108         if(now==t){109             maxflow+=augment(s,t);110             now=s;111         }112         bool ok=0;113         for(int p=cur[now];p;p=NE[p].next){114             j=NE[p].to;115             if(d[j]+1==d[now]&&NE[p].cap>NE[p].flow){116                 ok=1;117                 cur[now]=fl[j]=p;118                 now=j;119                 break;120             }121         }122         if(!ok){123             int minl=tot;124             for(int p=nfirst[now];p;p=NE[p].next){125                 j=NE[p].to;126                 if(d[j]+1<minl&&NE[p].cap>NE[p].flow) minl=d[j]+1;127             }128             if(--gap[d[now]]==0) break;129             gap[d[now]=minl]++;130             cur[now]=nfirst[now];131             if(now!=s) now=NE[fl[now]].from;132         }133     }134     return maxflow;135 }136 void work()137 {138     build_net();139     LL sum=0;140     for(int i=1;i<=N;i++) sum+=A[i],sum+=B[i];141     for(int i=1;i<=M;i++)142         sum+=C[i].a,sum+=C[i].b,sum+=C[i].c;143     cout<<sum-ISAP(S,T)/2<<'\n';144 }145 int main()146 {147     data_in();148     work();149     return 0;150 }