Hills And Valleys
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Special Judge
Problem Description
Tauren has an integer sequence A of length n (1-based). He wants you to invert an interval [l,r] (1≤l≤r≤n) of A (i.e. replace Al,Al+1,⋯,Ar with Ar,Ar−1,⋯,Al) to maximize the length of the longest non-decreasing subsequence of A. Find that maximal length and any inverting way to accomplish that mission.
A non-decreasing subsequence of A with length m could be represented as Ax1,Ax2,⋯,Axm with 1≤x1<x2<⋯<xm≤n and Ax1≤Ax2≤⋯≤Axm.
Input
The first line contains one integer T, indicating the number of test cases.
The following lines describe all the test cases. For each test case:
The first line contains one integer n.
The second line contains n integers A1,A2,⋯,An without any space.
1≤T≤100, 1≤n≤105, 0≤Ai≤9 (i=1,2,⋯,n).
It is guaranteed that the sum of n in all test cases does not exceed 2⋅105.
Output
For each test case, print three space-separated integers m,l and r in one line, where m indicates the maximal length and [l,r] indicates the relevant interval to invert.
Sample Input
2 9 864852302 9 203258468
Sample Output
5 1 8
6 1 2
题意：给你长度为n的字符串str[]，只包含'0'~'9'，你可以翻转一个区间，使得最后最长不下降子序列最长， 求出长度和你翻转的区间（输出任意一种即可）
假设没有翻转操作，答案就是 str[] 和 "0123456789" 的最长单字符可重公共子序列
单字符可重就是指的对于串 "0123456789" 中的每一个数字都可以匹配无数次再匹配下个数字，例如"001122223456789"等等
这道题可以翻转一次，那么可以暴力翻转区间中对答案有贡献的最小值和最大值[x, y]
假设对于当前其中一种情况[2, 3]，只需要求出 str[] 和 "01232345678" 的最长单字符可重公共子序列即可
对于输出翻转区间记录路径就好了
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[100005];
int tx, ty, a[100005], pre[100005][15], dp[100005][15], now[15], x1[100005][15], x2[100005][15];
int main(void)
{
int T, n, i, j, p, q, ans, cnt, ax, ay, Ans, c1, c2;
scanf("%d", &T);
while(T--)
{
tx = ty = 1;
scanf("%d%s", &n, str+1);
for(i=1;i<=n;i++)
a[i] = str[i]-'0';
for(i=1;i<=n;i++)
{
for(j=0;j<=9;j++)
pre[i][j] = pre[i-1][j];
pre[i][a[i]] = i;
}
Ans = ans = 0;
for(p=0;p<=9;p++)
{
for(q=p;q<=9;q++)
{
cnt = 0;
for(i=0;i<=p;i++) now[++cnt] = i;
for(i=q;i>=p;i--) now[++cnt] = i;
for(i=q;i<=9;i++) now[++cnt] = i;
for(i=1;i<=n;i++)
{
for(j=1;j<=cnt;j++)
{
if(a[i]==now[j])
{
dp[i][j] = dp[i-1][j-1]+1;
x1[i][j] = i-1, x2[i][j] = j-1;
if(dp[i-1][j]+1>dp[i][j])
{
dp[i][j] = dp[i-1][j]+1;
x2[i][j] = j;
}
}
else
{
dp[i][j] = dp[i-1][j];
x1[i][j] = i-1, x2[i][j] = j;
if(dp[i][j-1]>dp[i][j])
{
dp[i][j] = dp[i][j-1];
x1[i][j] = i, x2[i][j] = j-1;
}
}
if(dp[i][j]>=ans)
{
ans = dp[i][j];
ax = i, ay = j;
}
}
}
if(ans>Ans)
{
Ans = ans;
while(ax!=0)
{
if(ay==p+2 && x2[ax][ay]!=ay)
tx = ax;
if(ay==q+3 && x2[ax][ay]!=ay)
ty = x1[ax][ay];
c1 = x1[ax][ay];
c2 = x2[ax][ay];
ax = c1, ay = c2;
}
}
}
}
if(ty==0)
tx = ty = 1;
printf("%d %d %d\n", ans, tx, ty);
}
return 0;
}
/*
1
2
10
*/