Gym - 100548C The Problem Needs 3D Arrays

Problem C.   The Problem Needs 3D Arrays

　　Time Limit: 6000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

Description

A permutation is a sequence of integers p₁,p₂,...,p_n, consisting of n distinct positive integers and each of them does not exceed n. Assume that r(S) of sequence S denotes the number of inversions in sequence S (if i < j and S_i > S_j, then the pair of (i,j) is called an inversion of S), l(S) of sequence S denotes the length of sequence S. Given a permutation P of length n, it’s your task to find a subsequence S of P with maximum. A subsequence of P is a sequence (p_i1,p_i2,...,p_it) which satisfies that 0 < i₁ < i₂ < ... < i_t ≤ n.

Input

The first line of the 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 of the permutation P. The second line contains n integers p₁,p₂,...,p_n, which represents the permutation P.

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the maximum.

Your answer will be considered correct if it is within an absolute error of 10−⁶ of the correct answer.

Samples

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

解题：最大密度子图转为最大权闭合图

 

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 const int maxn = 12000;

 4 const int INF = 0x3f3f3f3f;

 5 int n,m,tot,S,T,x[maxn],y[maxn],A[102];

 6 struct arc {

 7     int to,next;

 8     double flow;

 9     arc(int x = 0,double y = 0,int z = -1) {

10         to = x;

11         flow = y;

12         next = z;

13     }

14 } e[maxn<<4];

15 int head[maxn],cur[maxn],d[maxn];

16 void add(int u,int v,double flow) {

17     e[tot] = arc(v,flow,head[u]);

18     head[u] = tot++;

19     e[tot] = arc(u,0,head[v]);

20     head[v] = tot++;

21 }

22 queue<int>q;

23 bool bfs() {

24     memset(d,-1,sizeof d);

25     while(!q.empty()) q.pop();

26     q.push(S);

27     d[S] = 1;

28     while(!q.empty()) {

29         int u = q.front();

30         q.pop();

31         for(int i = head[u]; ~i; i = e[i].next) {

32             if(e[i].flow > 0 && d[e[i].to] == -1) {

33                 d[e[i].to] = d[u] + 1;

34                 q.push(e[i].to);

35             }

36         }

37     }

38     return d[T] > -1;

39 }

40 double dfs(int u,double low) {

41     if(u == T) return low;

42     double tmp = 0,a;

43     for(int &i = cur[u]; ~i; i = e[i].next) {

44         if(e[i].flow > 0 && d[e[i].to] == d[u] + 1 &&(a=dfs(e[i].to,min(e[i].flow,low)))>0) {

45             e[i].flow -= a;

46             e[i^1].flow += a;

47             low -= a;

48             tmp += a;

49             if(low <= 0) break;

50         }

51     }

52     if(tmp <= 0) d[u] = -1;

53     return tmp;

54 }

55 bool dinic() {

56     double ans = m;

57     while(bfs()) {

58         memcpy(cur,head,sizeof head);

59         ans -= dfs(S,INF);

60     }

61     return ans <= 0;

62 }

63 void build(double delta) {

64     memset(head,-1,sizeof head);

65     for(int i = tot = 0; i < m; ++i) {

66         add(S,i + n + 1,1.0);

67         add(i + n + 1,x[i],INF);

68         add(i + n + 1,y[i],INF);

69     }

70     for(int i = 1; i <= n; ++i) add(i,T,delta);

71 }

72 int main() {

73     int cm = 1,cs;

74     scanf("%d",&cs);

75     while(cs--) {

76         scanf("%d",&n);

77         m = 0;

78         for(int i = 1; i <= n; ++i) {

79             scanf("%d",A+i);

80             for(int j = i-1; j > 0; --j)

81                 if(A[j] > A[i]) {

82                     x[m] = j;

83                     y[m++] = i;

84                 }

85         }

86         S = 0;

87         T = n + m + 1;

88         double low = 0,high = m,ans = 0;

89         if(m == 0) {printf("Case #%d: %.7f\n",cm++,ans);continue;}

90         while(high - low > 1e-7){

91           double mid = (low + high)/2.0;

92           build(mid);

93           if(dinic()) ans = high = mid;

94           else low = mid;

95         }

96         printf("Case #%d: %.7f\n",cm++,ans);

97     }

98     return 0;

99 }

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