2014 西安 The Problem Needs 3D Arrays

The Problem Needs 3D Arrays

题意：给你n个数， 然后1-n的数， 然后要求按顺序选出m个数， 求 逆序数/m 个数的 最大值是多少。

题解：裸的最大密度子图。逆序的2个数建边， 跑一下最大密度子图就AC了。

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const int inf = 0x3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = ;
 const int M = ;
 double eps = 1e-;
 int head[N], deep[N], cur[N];
 int u[M], v[M];
 int vis[N], d[N];
 double w[M]; int to[M], nx[M];
 int n, m, tot;
 int a[N];
 void add(int u, int v,double val){
     w[tot]  = val; to[tot] = v;
     nx[tot] = head[u]; head[u] = tot++;
     w[tot]  = ; to[tot] = u;
     nx[tot] = head[v]; head[v] = tot++;
 }
 int bfs(int s, int t){
     queue<int> q;
     memset(deep, , sizeof(int)*(n+));
     q.push(s);
     deep[s] = ;
     while(!q.empty()){
         int u = q.front();
         q.pop();
         for(int i = head[u]; ~i; i = nx[i]){
             if(w[i] >  && deep[to[i]] == ){
                 deep[to[i]] = deep[u] + ;
                 q.push(to[i]);
             }
         }
     }
     if(deep[t] > ) return ;
     return ;
 }
 double Dfs(int u, int t, double flow){
     if(u == t) return flow;
     for(int &i = cur[u]; ~i; i = nx[i]){
         if(deep[u] +  == deep[to[i]] && w[i] > ){
             double di = Dfs(to[i], t, min(w[i], flow));
             if(di > ){
                 w[i] -= di, w[i^] += di;
                 return di;
             }
         }
     }
     return ;
 }

 int Dinic(int s, int t){
     double ans = , tmp;
     while(bfs(s, t)){
         for(int i = ; i <= n+; i++) cur[i] = head[i];
         while(tmp = Dfs(s, t, INF)) ans += tmp;
     }
     return ans;
 }

 bool check(double x){
     memset(head, -, sizeof(int) * (n+));
     tot = ;
     int s = , t = n + ;
     for(int i = ; i <= m; i++){
         add(u[i], v[i], );
         add(v[i], u[i], );
     }
     for(int i = ; i <= n; i++){
         add(s, i, m);
         add(i, t, m+*x-d[i]);
     }
     return (m*n-Dinic(s,t))/2.0 >= 1e-;
 }

 int main(){
     int T;
     scanf("%d", &T);
     for(int _i = ; _i <= T; _i++){
         scanf("%d", &n);
         for(int i = ; i <= n; i++)scanf("%d", &a[i]);
         memset(d, , sizeof(int)*(n+));
         m = ;
         for(int i = ; i <= n; i++){
             for(int j = ; j < i; j++){
                 if(a[j] > a[i]){
                     d[i]++;
                     d[j]++;
                     m++;
                     u[m] = i;
                     v[m] = j;
                 }
             }
         }
         double l = , r = m,  mid;
         while(r - l >= eps){
             mid = (l+r)/;
             if(check(mid)) l = mid;
             else r = mid;
         }
         printf("Case #%d: %.12f\n", _i, l);
         //printf
     }

     return ;
 }