CF GYM 100548 The Problem Needs 3D Arrays（2014ACM西安现场赛Problem C）

ProblemC. The Problem Needs 3D Arrays

Description
A permutation is asequence of integers p1, p2, . . . , pn,consisting of n distinct positive integers and each of them does notexceed n. Assume that r(S) of sequence S denotes the
number ofinversions in sequence S (if i < j and Si > Sj,then the pair of (i, j) is called an inversion of S), l(S) ofsequence S denotes the length of sequence S. Given a permutation P oflength n, it’s your task to find a subsequence S
of P with maximumr(S) / l(S). A subsequence of P is a sequence (pi1, pi2,. . . , pit) which satisfies that 0 < i1 <i2 < . . . < it ≤ n.

Input
The first line ofthe input gives the number of test cases, T. T test cases follow.
For each test case,the first line contains an integer n (1 ≤ n ≤ 100), the length ofthe permutation P. The second line contains n integers p1,p2, . . . , pn, which represents thepermutation
P.

Output
For each test case,output one line containing “Case #x: y”, where x is the test casenumber (starting from 1) and y is the maximum r(S) / l(S).
Your answer will beconsidered correct if it is within an absolute error of 10−6of the correct answer.

Samples

Sample Input	Sample Output
1 5 3 4 2 5 1	Case #1: 1.250000000000

知识点：
最大密度子图、最大权闭合图、最小割。

题目大意：
给出1～n这n个正整数的一种排列P，要求在P中找出一个子序列S，使得子序列中包含的逆序数对数r(S)和子序列的长度l(S)的比值最大。输出这个最大的比值r(S)/
l(S)。

解题思路：

可以将每个数看成图中的点，将逆序对的关系转换为图中的边。即构成了一个无向图。样例可以转换为下图：

现在要求的就是在图中选取一些点，以及他们互相之间相连的边，构成一个闭合子图。使得图中的边数（逆序对数）与点数（选取的数字）的比值最大。这就是一个最大密度子图的模型。样例的最佳选取方案，就是选择图中红色标注的点和边，共4个点5条边，比值为1.25。
根据胡伯涛的论文《最小割模型在信息学竞赛中的应用》，可以使用0-1分数规划的模型。根据分数规划一般步骤，二分查找答案。对于一个答案的猜测值g，将原图转化为网络G(V,E)的过程，即在原图点集V的基础上增加源s和汇t；将每条原无向边替换为两条容量为1的有向边(u,v)和(v,u)；增加连接源到原图每个点的有向边(s,v)，容量为U；增加连接原图每个点v到汇t的有向边(v,t)，容量为(U+2g−d[v])（其中d[v]是v点的度数）。求出网络的最小割（等于最大流），如果(U×n-最小割)/
2 >=0，则猜测值g是可行的，放大g，继续二分；若<0则是不可行的，缩小g，继续二分。最后得出的就是可行的最大的g。即可求出最大的比值。

参考代码：（未AC）

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;

const double INF = 1 << 30;
const double EPS = 1e-12;
const int MAXN = 110;
const int MAXM = 50010;

struct Edge {
int u, v, next;
double c;
Edge() {}
Edge(int _u, int _v, double _c, int _next) : u(_u), v(_v), c(_c), next(_next) {}
} edge[MAXM];

int head[MAXN], cnt;
bool visited[MAXN];
int path[MAXN], d[MAXN], src, to;
int n, m, nCase, cCase, p[MAXN], degree[MAXN];
vector<pair<int, int> > E;

void addEdge(int u, int v, double c) {
edge[cnt] = Edge(u, v, c, head[u]);
head[u] = cnt++;
edge[cnt] = Edge(v, u, 0, head[v]);
head[v] = cnt++;
}

bool bfs() {
memset(d, 0, sizeof(d));
queue<int> q;
q.push(src);
d[src] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].v;
if (edge[i].c && !d[v]) {
d[v] = d[u] + 1;
q.push(v);
}
}
}
return d[to];
}

double dfs(int x, double pf) {
if (x == to) return pf;
double ret = 0;
for (int i = head[x]; i != -1 && pf; i = edge[i].next) {
int v = edge[i].v;
if (edge[i].c && d[v] == d[x] + 1) {
double p = dfs(v, min(edge[i].c, pf));
edge[i].c -= p;
edge[i ^ 1].c += p;
ret += p;
pf -= p;
}
}
if (!ret) d[x] = -2;
return ret;
}

double dinic() {
double ret = 0.0;
while (bfs()) ret += dfs(src, INF);
return ret;
}

inline void buildEdge(double g) {
memset(head, -1, sizeof(head));
cnt = 0;

src = 0;
to = n + 1;

for (int i = 0; i < m; i++) {
addEdge(E[i].first, E[i].second, 1.0);
addEdge(E[i].second, E[i].first, 1.0);
}
for (int i = 1; i <= n; i++) {
addEdge(src, i, m * 1.0);
addEdge(i, to, m * 1.0 + 2.0 * g - degree[i]);
}
}

void init() {
E.clear();
memset(degree, 0, sizeof(degree));
}

void input() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &p[i]);
}
}

inline bool check(double g) {
buildEdge(g);
double h = (1.0 * m * n - dinic()) / 2.0;
return h > 0;
}

void solve() {
for (int i = 1; i <= n; i++) {
for (int j = i+1; j <= n; j++) {
if (p[i] > p[j]) {
E.push_back(make_pair(i, j));
degree[i]++;
degree[j]++;
}
}
}
m = E.size();

if (m == 0) {
printf("Case #%d: %.12lf\n", ++cCase, 0);
return;
}

double l = 1.0 / n - EPS, r = m * 1.0 + EPS;
while (r - l > EPS) {
double mid = (l + r) / 2.0;
if (check(mid)) {
l = mid;
} else {
r = mid;
}
}

printf("Case #%d: %.12lf\n", ++cCase, l);
}

int main() {
scanf("%d", &nCase);
while (nCase--) {
init();
input();
solve();
}
return 0;
}

欢迎讨论指正。