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将A[i]同他后面比他小的建边,然后求最大密度子图
#include <iostream> #include <algorithm> #include <string.h> #include <cstdio> #include <vector> #include <queue> #include <cmath> using namespace std; const int maxn=105; const double eps=0.00000001; const double INF=100000005; vector<int>ans; struct Dinic { struct Edge { int from,to; double cap,flow; Edge(int cfrom=0,int cto=0,double ccap=0,double cflow=0) { from=cfrom; to=cto; cap=ccap; flow=cflow; } }; int n,m,s,t; vector<Edge>edges; vector<int>G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; void init(int n) { this->n=n; m=0; edges.clear(); for(int i=0; i<=n; i++)G[i].clear(); } void AddEdge(int from,int to,double cap) { edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); m+=2; G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS() { memset(vis,false,sizeof(vis)); queue<int>Q; Q.push(s); d[s]=0; vis[s]=1; while(!Q.empty()) { int x=Q.front(); Q.pop(); for(int i=0; i<G[x].size(); i++) { Edge &e =edges[G[x][i]]; if(vis[e.to]==false&&e.cap>e.flow) { vis[e.to]=1; d[e.to]=d[x]+1; Q.push(e.to); } } } return vis[t]; } double DFS(int x, double a) { if(x==t||a==0)return a; double flow=0,f; for(int &i=cur[x]; i<G[x].size(); i++) { Edge &e=edges[G[x][i]]; if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0) { e.flow+=f; edges[G[x][i]^1].flow-=f; flow+=f; a-=f; if(a==0)break; } } return flow; } double Maxflow(int s,int t) { this->s=s; this->t=t; double flow=0; while(BFS()) { memset(cur,0,sizeof(cur)); flow+=DFS(s,INF); } return flow; } }T; int d[maxn]; int A[maxn*maxn],B[maxn*maxn]; int N[maxn]; double U; void build(int n,int m,double g) { T.init(n+2); for(int i=1; i<=n; i++) { T.AddEdge(n+1,i,U); T.AddEdge(i,n+2,U+g*2-d[i]); } for(int i=0; i<m; i++) { T.AddEdge(A[i],B[i],1); T.AddEdge(B[i],A[i],1); } } int main() { int cas; scanf("%d",&cas); for(int cc=1; cc<=cas; cc++) { int n; scanf("%d",&n); for(int i=1; i<=n; i++) scanf("%d",&N[i]); int m=0; memset(d,0,sizeof(d)); for(int i=1; i<=n; i++) for(int j=i+1; j<=n; j++) if(N[i]>N[j]){ d[N[i]]++;d[N[j]]++; A[m]=N[i];B[m]=N[j]; m++; } if(m==0) { printf("Case #%d: 0.0000000\n",cc); continue; } double L=0,R=m; U=m; while(R-L>=eps) { double mid=(L+R)*0.5; build(n,m,mid); double ans=T.Maxflow(n+1,n+2); ans=(U*n-ans)*0.5; if(ans>0)L=mid; else R=mid; } printf("Case #%d: %.7lf\n",cc,L ); } return 0; } /* 2 5 3 4 2 5 1 5 2 1 3 4 5 */
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原文地址:http://www.cnblogs.com/Opaser/p/4922061.html