NEFU 708 Longest Non-decreasing Substring(线段树+DP)

Longest Non-decreasing Substring

Time Limit 1000ms

Memory Limit 65536K

description

You are given a string S no longer than 100,000 consisting of characters 0-9 only. And two kinds of operations will be performed on S:
Flip i j: for each i <= k <= j, Sk is replaced substituted with 9 - Sk.
Query: you are to calculate the length of the longest non-decreasing substring of S.
							

input

The first line of the input contains a single integer t, the number of test cases.
For each test case, the first line contains two integers, n and m(1 <= m <= 5000), indicating the length of S and the number of operations to perform.
Then m lines follow. Each line gives a single operation that is either “flip i j”(1 <= i <= j <= n) or “query”.
There is a blank line after each test case.
							

output

For each query operation, print the length of the longest non-decreasing substring of S on a single line.
Print a blank line after each test case.
							

sample_input

1
10 6
0123456789
query
flip 1 10
query
flip 1 10
flip 3 4
query
							

sample_output

10
1
6
							

hint

source

思路：线段树一个节点记录9个值，de(区间的最长递减序)inc(区间的最长递增)lz(区间最左值)rz(区间最右值)ld(从最左开始区间连续递减数)rd(从区间最右开始区间连续递减数)

          lg(区间最左开始连续递增数)rg(区间最右开始连续递增数)lazy（懒惰标记）

        翻转更新：把该区间的的最左值，最右值9- ，递增与递减交换。

        更新值时，应注意中间段连续需要连接。

失误点：lazy传完忘记赋成0了

对于这样看起来能做但是比赛没码出来，比完后调出来的的题，真心感觉很痛苦，这已经是第几题这样了？？？是水平仍旧不够，还是没有状态？？？哎

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define lson t<<1
#define rson t<<1|1
#define midl (l+r)/2
#define midr (l+r)/2+1
using namespace std;
const int mm=1e5+9;
class node
{
  public:int l,r,de,inc,lz,rz,ld,rd,lg,rg,lazy;
}rt[mm*4];
char s[mm];

void chage(int t)///翻转
{
  ++rt[t].lazy;
  swap(rt[t].de,rt[t].inc);
  swap(rt[t].ld,rt[t].lg);
  swap(rt[t].rd,rt[t].rg);
  rt[t].lz=9-rt[t].lz;rt[t].rz=9-rt[t].rz;
}
void upush(int t)///更新值
{
  rt[t].lz=rt[lson].lz;rt[t].rz=rt[rson].rz;
  rt[t].lg=rt[lson].lg;rt[t].rg=rt[rson].rg;
  rt[t].ld=rt[lson].ld;rt[t].rd=rt[rson].rd;
  rt[t].inc=max(rt[lson].inc,rt[rson].inc);
  rt[t].de=max(rt[lson].de,rt[rson].de);
  if(rt[lson].rz<=rt[rson].lz)///中间可递增
    {rt[t].inc=max(rt[t].inc,rt[lson].rg+rt[rson].lg);
     if(rt[t].lg==rt[lson].r-rt[lson].l+1)rt[t].lg+=rt[rson].lg;
     if(rt[t].rg==rt[rson].r-rt[rson].l+1)rt[t].rg+=rt[lson].rg;
    }
  if(rt[lson].rz>=rt[rson].lz)///中间可递减
    {rt[t].de=max(rt[t].de,rt[lson].rd+rt[rson].ld);
     if(rt[t].ld==rt[lson].r-rt[lson].l+1)rt[t].ld+=rt[rson].ld;
     if(rt[t].rd==rt[rson].r-rt[rson].l+1)rt[t].rd+=rt[lson].rd;
    }
}
void build(int t,int l,int r)
{
  rt[t].l=l;rt[t].r=r;rt[t].lazy=0;
  if(l==r)
    {
      rt[t].lz=rt[t].rz=s[r-1]-'0';
      rt[t].de=rt[t].inc=rt[t].ld=rt[t].rd=rt[t].lg=rt[t].rg=1;
      return;
    }
  build(lson,l,midl);build(rson,midr,r);
  upush(t);
}
void update(int t,int l,int r)
{
  if(rt[t].l>=l&&rt[t].r<=r)
  { /**swap(rt[t].de,rt[t].inc);
    swap(rt[t].ld,rt[t].lg);
    swap(rt[t].rd,rt[t].rg);
    ++rt[t].lazy;*/
    chage(t);
    return;
    ///rt[t].de^=rt[t].inc;rt[t].inc^=rt[t].de;rt[t].de^=rt[t].inc;
  }
  if(rt[t].lazy&1)
  {
    chage(lson);chage(rson);rt[t].lazy=0;
  }
  if(l<=rt[lson].r)update(lson,l,r);
  if(r>=rt[rson].l)update(rson,l,r);
 /// else update(lson,l,r),update(rson,l,r);
  upush(t);
}
int main()
{
  int cas,n,m,a,b;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    {
      scanf("%d%d",&n,&m);
      scanf("%s",s);
      n=strlen(s);
      build(1,1,n);
      for(int i=0;i<m;++i)
      {
        scanf("%s",s);
        if(s[0]=='q')printf("%d\n",rt[1].inc);
        else
        { scanf("%d%d",&a,&b);
          update(1,a,b);
        }
      }
      printf("\n");
    }
  }
}