最大流 ---- 最大密度子图 ----- 2014-2015 ACM-ICPC, Asia Xian Regional Contest C The Problem Needs 3D Arrays

题目链接

题目大意：

就是给你一个全排列，现在叫你选出一个子序列sss，使得r(s)l(s)\frac{r(s)}{l(s)}l(s)r(s)最大？

r(s)r(s)r(s)是这个子序列的逆序对数
l(s)l(s)l(s)是这个子序列的长度
n∈[1,100]n\in[1,100]n∈[1,100]

看到这个分数和n这么小，就不自觉的联想到最大密度子图，但是图是怎么样的呢？

图就是对于任意的i<j&&ai>aji<j\&\&a_i>a_ji<j&&ai>aj那么i和j之间就有一条边i和j之间就有一条边i和j之间就有一条边
那么直接跑最大密度子图就好了

#include <bits/stdc++.h>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define lowbit(x) ((-x) & x)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define f first
#define s second
#define endl '\n'
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 100010;
const double eps = 1e-7;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
struct node {int to, next;double len;
}e[maxn];
int head[maxn], cnt;
int n, m, s, t;inline void add(int from, int to, double len) {e[cnt] = {to,head[from],len};head[from] = cnt ++;
}int d[maxn],cur[maxn],gap[maxn];int arr[maxn];void bfs()//最短路
{ms(d,0);ms(gap,0);queue<int> q;q.push(t); d[t] = gap[1] = 1;while(!q.empty()) {int u = q.front();q.pop();for(int i = head[u]; ~i; i = e[i].next){int v = e[i].to;if(d[v]) continue;q.push(v); d[v] = d[u] + 1;++ gap[d[v]];}}_for(i,0,t+2) cur[i] = head[i];
}double dfs(int u, double flow)
{if(u == t) return flow;double delta = 0, tmp;for(int &i = cur[u]; ~i; i = e[i].next){int v = e[i].to;if(d[u] - 1 != d[v] || e[i].len <= 0) continue;double tmp = dfs(v,min(flow - delta,e[i].len));if(tmp <= eps) continue;e[i].len -= tmp;e[i^1].len += tmp;delta += tmp;if(fabs(delta - flow) < eps) return delta;}--gap[d[u]];if(!gap[d[u]]) d[s] = t + 1;++ gap[++ d[u]];cur[u] = head[u];return delta;
}double get_maxflow() {bfs();double maxFlow = 0;while(d[s] <= t)maxFlow += dfs(s,1e9);return maxFlow;
}PII edge[maxn];inline void build(double Mid) {ms(head,-1);cnt = 0;s = 0, t = n + m + 1;for(int i = 1; i <= m; ++ i) {int u = edge[i].first, v = edge[i].second;add(s,i+n,1);add(i+n,s,0);add(i+n,u,1e9);add(u,i+n,0);add(i+n,v,1e9);add(v,i+n,0);}for(int i = 1; i <= n; ++ i) {add(i,t,Mid);add(t,i,0);}}int main() {IOS;int T;cin >> T;for(int Cas = 1; Cas <= T; ++ Cas) {cin >> n;m = 0;for(int i = 1; i <= n; ++ i) cin >> arr[i];for(int i = 1; i <= n; ++ i) for(int j = i + 1; j <= n; ++ j) if(arr[i] > arr[j]) edge[++ m] = {i,j}; cout << "Case #" << Cas << ": ";if(!m)  cout << "0.0000000\n";else {double l = 1.0/n, r = m;while(r - l > eps) {double Mid = (l + r) / 2;build(Mid);if(1.0 * m - get_maxflow() > eps) l = Mid;else r = Mid;}cout << fixed << setprecision(10) << l << '\n';}}return 0;
}

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最大流 ---- 最大密度子图 ----- 2014-2015 ACM-ICPC, Asia Xian Regional Contest C The Problem Needs 3D Arrays

题目大意：

最大流 ---- 最大密度子图 ----- 2014-2015 ACM-ICPC, Asia Xian Regional Contest C The Problem Needs 3D Arrays相关推荐

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