C. An impassioned circulation of affection
2017-06-08 16:50
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C. An impassioned circulation of affection
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from
left to right, and the i-th piece has a colour si,
denoted by a lowercase English letter. Nadeko will repaint at most m of
the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c —
Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o".
Among all subsegments containing pieces of "o" only, "ooo"
is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans
on this, each of which can be expressed as a pair of an integer mi and
a lowercase letter ci,
meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.
Input
The first line of input contains a positive integer n (1 ≤ n ≤ 1 500)
— the length of the garland.
The second line contains n lowercase English letters s1s2... sn as
a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer q (1 ≤ q ≤ 200 000)
— the number of plans Nadeko has.
The next q lines describe one plan each: the i-th
among them contains an integer mi (1 ≤ mi ≤ n)
— the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci —
Koyomi's possible favourite colour.
Output
Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable
after repainting the garland according to it.
Examples
input
output
input
output
input
output
Note
In the first sample, there are three plans:
In the first plan, at most 1 piece can be repainted. Repainting the "y"
piece to become "o" results in "kooomi", whose Koyomity of 3is
the best achievable;
In the second plan, at most 4 pieces can be repainted, and "oooooo"
results in a Koyomity of 6;
In the third plan, at most 4 pieces can be repainted, and "mmmmmi"
and "kmmmmm" both result in a Koyomity of 5.
不得不说二分真是一个神奇的东西,嗯应该说是优化利器。
昨天做的时候一直T,也灭有想到什么好的优化方法,没错,还有这种操作,二分。
枚举左区间,然后二分右区间,复杂度瞬间从n3变成nlogn。
QAQ 又掉分了,太惨了。。。
没有预处理,直接记忆化了一下。
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1500+7;
const int inf = 1e9;
int n,m;
char s[MAXN];
int pos[30][MAXN];
int temp[MAXN];
int ans[1501][30];
int get_ans(int x,char c)
{
int x1 = c - 'a';
for(int i = 1 ; i <= n ; ++i)
{
temp[i] = temp[i-1] + pos[x1][i];
}
int MAX = x;
for(int i = 1 ; i < n ; ++i)
{
int low = i,high = n,ans;
while(low <= high)
{
int mid = (low + high)>>1;
int l = mid - i + 1;
if(l - (temp[mid] - temp[i-1]) <= x)
{
ans = mid;
low = mid + 1;
}
else high = mid - 1;
}
MAX = max(MAX, ans - i + 1);
}
return MAX;
}
int main()
{
scanf("%d",&n);
scanf("%s",s+1);
int l = strlen(s+1);
for(int i = 1 ; i <= l ; ++i)
{
int x1 = s[i] - 'a';
pos[x1][i]++;
}
scanf("%d",&m);
int x;
char c[2];
while(m--)
{
scanf("%d%s",&x,c);
int x1 = c[0] - 'a';
if(!ans[x][x1])ans[x][x1] = get_ans(x,c[0]);
printf("%d\n",ans[x][x1]);
}
}
貌似人家的dp要快很多。。。
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from
left to right, and the i-th piece has a colour si,
denoted by a lowercase English letter. Nadeko will repaint at most m of
the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c —
Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o".
Among all subsegments containing pieces of "o" only, "ooo"
is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans
on this, each of which can be expressed as a pair of an integer mi and
a lowercase letter ci,
meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.
Input
The first line of input contains a positive integer n (1 ≤ n ≤ 1 500)
— the length of the garland.
The second line contains n lowercase English letters s1s2... sn as
a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer q (1 ≤ q ≤ 200 000)
— the number of plans Nadeko has.
The next q lines describe one plan each: the i-th
among them contains an integer mi (1 ≤ mi ≤ n)
— the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci —
Koyomi's possible favourite colour.
Output
Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable
after repainting the garland according to it.
Examples
input
6 koyomi 3 1 o 4 o 4 m
output
3 6 5
input
15 yamatonadeshiko 10 1 a 2 a 3 a 4 a 5 a 1 b 2 b 3 b 4 b 5 b
output
3 4 5 7 8 1 2 3 4 5
input
10 aaaaaaaaaa 2 10 b 10 z
output
10 10
Note
In the first sample, there are three plans:
In the first plan, at most 1 piece can be repainted. Repainting the "y"
piece to become "o" results in "kooomi", whose Koyomity of 3is
the best achievable;
In the second plan, at most 4 pieces can be repainted, and "oooooo"
results in a Koyomity of 6;
In the third plan, at most 4 pieces can be repainted, and "mmmmmi"
and "kmmmmm" both result in a Koyomity of 5.
不得不说二分真是一个神奇的东西,嗯应该说是优化利器。
昨天做的时候一直T,也灭有想到什么好的优化方法,没错,还有这种操作,二分。
枚举左区间,然后二分右区间,复杂度瞬间从n3变成nlogn。
QAQ 又掉分了,太惨了。。。
没有预处理,直接记忆化了一下。
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1500+7;
const int inf = 1e9;
int n,m;
char s[MAXN];
int pos[30][MAXN];
int temp[MAXN];
int ans[1501][30];
int get_ans(int x,char c)
{
int x1 = c - 'a';
for(int i = 1 ; i <= n ; ++i)
{
temp[i] = temp[i-1] + pos[x1][i];
}
int MAX = x;
for(int i = 1 ; i < n ; ++i)
{
int low = i,high = n,ans;
while(low <= high)
{
int mid = (low + high)>>1;
int l = mid - i + 1;
if(l - (temp[mid] - temp[i-1]) <= x)
{
ans = mid;
low = mid + 1;
}
else high = mid - 1;
}
MAX = max(MAX, ans - i + 1);
}
return MAX;
}
int main()
{
scanf("%d",&n);
scanf("%s",s+1);
int l = strlen(s+1);
for(int i = 1 ; i <= l ; ++i)
{
int x1 = s[i] - 'a';
pos[x1][i]++;
}
scanf("%d",&m);
int x;
char c[2];
while(m--)
{
scanf("%d%s",&x,c);
int x1 = c[0] - 'a';
if(!ans[x][x1])ans[x][x1] = get_ans(x,c[0]);
printf("%d\n",ans[x][x1]);
}
}
貌似人家的dp要快很多。。。
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