杭电2019多校第一场，Problem I，String 2019

题目描述

Tom has a string containing only lowercase letters. He wants to choose a subsequence of the string whose length is k and lexicographical order is the smallest. It's simple and he solved it with ease.
But Jerry, who likes to play with Tom, tells him that if he is able to find a lexicographically smallest subsequence satisfying following 26 constraints, he will not cause Tom trouble any more.
The constraints are: the number of occurrences of the ith letter from a to z (indexed from 1 to 26) must in [Li,Ri].
Tom gets dizzy, so he asks you for help.
 

 

输入

The input contains multiple test cases. Process until the end of file.
Each test case starts with a single line containing a string S(|S|≤105)and an integer k(1≤k≤|S|).
Then 26 lines follow, each line two numbers Li,Ri(0≤Li≤Ri≤|S|). 
It's guaranteed that S consists of only lowercase letters, and ∑|S|≤3×105.

 

输出

Output the answer string.
If it doesn't exist, output −1.

样例输入

aaabbb 3
0 3
2 3
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0

样例输出

abb

题解：
将字母下标按字母从前往后依次抽取出来；
然后从前往后依次确定k的每一位应该放的字母；
对于每一位，枚举字母的顺序应该是从a到z，接着判断该字母放上去后是否符合条件，符合则去确定k的下一位字母，不符合则继续循环；
时间复杂度应该是O(26*26*n)。
AC代码：
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+50;
 4 int suf_sum[maxn][30],l[30],r[30],used[30],n,k,last;
 5 char str[maxn],res[maxn];
 6 vector<int> letter_index[30];
 7 int main()
 8 {
 9     while(scanf("%s %d",str,&k)!=EOF)
10     {
11         for(int i=0;i<=25;i++){
12             scanf("%d %d",&l[i],&r[i]);
13             letter_index[i].clear();
14         }
15         vector<int> ::iterator head[30];
16         memset(used,0,sizeof(used));
17         n=strlen(str);last=-1;
18         for(int i=0;i<=n;i++) for(int j=0;j<=25;j++) suf_sum[i][j]=0;
19         for(int i=n-1;i>=0;i--){
20             for(int j=0;j<=25;j++){
21                 if(str[i]=='a'+j) suf_sum[i][j]=suf_sum[i+1][j]+1;
22                 else suf_sum[i][j]=suf_sum[i+1][j];
23             }
24         }
25         for(int i=0;i<=n-1;i++){
26             letter_index[str[i]-'a'].push_back(i);
27         }
28         for(int i=0;i<=25;i++){
29             head[i]=letter_index[i].begin();
30         }
31         bool ans=true;
32         for(int i=0;i<=k-1;i++){
33             bool flag=false;
34             for(int j=0;j<=25;j++){
35                 if(used[j]==r[j]) continue;
36                 while(head[j]!=letter_index[j].end() && (*head[j])<=last) head[j]++;
37                 if(head[j]==letter_index[j].end()) continue;
38                 used[j]++;
39                 bool tag=true;
40                 int cnt=0,tmp=0,pos=(*head[j]);
41                 for(int t=0;t<=25;t++){
42                     if(suf_sum[pos+1][t]+used[t]<l[t]) tag=false;
43                     cnt+=max(0,l[t]-used[t]);
44                     tmp+=min(suf_sum[pos+1][t],r[t]-used[t]);
45                 }
46                 if(cnt>k-1-i || tmp<k-1-i) tag=false;
47                 if(!tag) used[j]--;
48                 else{
49                     res[i]='a'+j;
50                     last=pos;
51                     flag=true;
52                     break;
53                 }
54             }
55             if(!flag){
56                 ans=false;
57                 printf("-1\n");
58                 break;
59             }
60         }
61         if(ans){
62             res[k]='\0';
63             printf("%s\n",res);
64         }
65     }
66     return 0;
67 }
68 /*
69 aaccddaa 6
70 2 4
71 0 0
72 2 2
73 0 2
74 0 0
75 0 0
76 0 0
77 0 0
78 0 0
79 0 0
80 0 0
81 0 0
82 0 0
83 0 0
84 0 0
85 0 0
86 0 0
87 0 0
88 0 0
89 0 0
90 0 0
91 0 0
92 0 0
93 0 0
94 0 0
95 0 0
96 */
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posted @ 
2019-07-26 14:30 
lglh 
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