ZOJ 2702 Unrhymable Rhymes

Unrhymable Rhymes

Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

 

Description

An amateur poet Willy is going to write his first abstract poem. Since abstract art does not give much care to the meaning of the poem, Willy is planning to impress listeners with unusual combinations of words. He prepared n lines of the future poem, but suddenly noticed that not all of them rhyme well.

Though abstractionist, Willy strongly respects canons of classic poetry. He is going to write the poem that would consist of quatrains. Each quatrain consists of two pairs of rhymed lines. Therefore there can be four types of quatrains, if we denote rhymed lines with the same letter, these types are AABB, ABAB, ABBA and AAAA.

Willy divided the lines he composed into groups, such that in each group any line rhymes with any other one. He assigned a unique integer number to each group and wrote the number of the group it belongs next to each line. Now he wants to drop some lines from the poem, so that it consisted of correctly rhymed quatrains. Of course, he does not want to change the order of the lines.

Help Willy to create the longest poem from his material.

 

Input

 

There are mutilple cases in the input file.

The first line of each case contains n --- the number of lines Willy has composed (1 <= n <= 4000 ). It is followed by n integer numbers denoting the rhyme groups that lines of the poem belong to. All numbers are positive and do not exceed 10⁹ .

There is an empty line after each case.

 

Output

 

On the first line of the output file print k --- the maximal number of quatrains Willy can make. After that print 4k numbers --- the lines that should form the poem.

There should be an empty line after each case.

 

Sample Input

 

15
1 2 3 1 2 1 2 3 3 2 1 1 3 2 2

3
1 2 3

 

Sample Output

 

3
1 2 4 5
7 8 9 10
11 12 14 15

0

 

　　解题：贪心+离散

　　

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define INF 0x3f3f3f3f
15 #define pii pair<int,int>
16 using namespace std;
17 const int maxn = 5100;
18 int p[maxn];
19 int d[maxn],n;
20 int ans[maxn],cnt,tot;
21 vector<int>g[maxn];
22 int main(){
23     while(~scanf("%d",&n)){
24         for(int i = 0; i < n; ++i){
25             scanf("%d",d+i);
26             p[i]= d[i];
27         }
28         for(int i = tot = 0; i < maxn; ++i) g[i].clear();
29         sort(d,d+n);
30         for(int i = cnt = 1; i < n; ++i){
31             if(d[i] == d[cnt-1]) continue;
32             d[cnt++] = d[i];
33         }
34         bool flag = false;
35         for(int i = 0; i < n; ++i){
36             int index = lower_bound(d,d+cnt,p[i]) - d;
37             g[index].push_back(i+1);
38             if(g[index].size() == 4){
39                 for(int j = 0,k = g[index].size(); j < k; ++j)
40                     ans[tot++] = g[index][j];
41                 for(int i = 0; i < cnt; ++i) g[i].clear();
42                 flag = true;
43             }
44             if(g[index].size() == 2){
45                 int k;
46                 for(k = 0; k < cnt; ++k){
47                     if(k == index) continue;
48                     if(g[k].size() >= 2) break;
49                 }
50                 if(k < cnt){
51                     flag = true;
52                     for(int j = 0; j < 2; ++j)
53                         ans[tot++] = g[k][j];
54                     for(int j = 0; j < 2; ++j)
55                         ans[tot++] = g[index][j];
56                     for(k = 0; k < n; ++k) g[k].clear();
57                 }
58             }
59         }
60         sort(ans,ans+tot);
61         if(flag){
62             printf("%d\n",tot>>2);
63             for(int i = 0; i < tot; i += 4)
64                 printf("%d %d %d %d\n",ans[i],ans[i+1],ans[i+2],ans[i+3]);
65         }else puts("0");
66         putchar('\n');
67     }
68     return 0;
69 }

View Code

 

转载于:https://www.cnblogs.com/crackpotisback/p/4069901.html