洛谷P3939 数颜色 二分查找

正解:二分

解题报告:

传送门!

话说其实我开始看到这题想到的是分块,,,

但是显然不用这么复杂,,,因为仔细看下这题,会发现每次只改变相邻的兔子的位置

所以开个vector(或者开个数组也成QwQ(数组就能用lower_bound

按顺序存下来每个颜色的兔子的位置,每次修改只用O(1)地改就好了

然后复杂度是O(nlogn),和莫队复杂度一样却简单很多

over

(对了,这题我本来想的是分块嘛,我就搜了下可不可以用分块,只看到了一篇题解,说分块会被时空双卡,但是可以优化,我还没有仔细看先贴个链接QAQ

然后放下二分代码QAQ

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define rp(i,x,y) for(register ll i=x;i<=y;++i)

const ll N=+;
ll n,m,a[N];
vector<ll>gg[N];

inline ll read()
{
    register char ch=getchar();register ll x=;register bool y=;
    while(ch!='-' && (ch>'' || ch<''))ch=getchar();
    if(ch=='-')ch=getchar(),y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=getchar();
    return y?x:-x;
}
inline void wk1(){ll l=read(),r=read(),c=read();if(gg[c].size()==){printf("0\n");return;}printf("%d\n",upper_bound(gg[c].begin(),gg[c].end(),r)-lower_bound(gg[c].begin(),gg[c].end(),l));}
inline void wk2(){ll x=read();if(a[x]==a[x+])return;gg[a[x]][lower_bound(gg[a[x]].begin(),gg[a[x]].end(),x)-gg[a[x]].begin()]=x+;gg[a[x+]][lower_bound(gg[a[x+]].begin(),gg[a[x+]].end(),x+)-gg[a[x+]].begin()]=x;swap(a[x],a[x+]);}

int main()
{
    n=read();m=read();rp(i,,n)a[i]=read(),gg[a[i]].push_back(i);
    while(m--){ll op=read();op==?wk1():wk2();}
    return ;
}

压行选手绝不轻易认输!(hhh其实我也jio得这次的代码压得丑了点儿,,,然而不压行我就,浑身难受QAQ

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define rp(i,x,y) for(register ll i=x;i<=y;++i)

const ll N=;
ll n,m,a[N];
vector<ll>gg[N];

inline ll read()
{
    register char ch=getchar();register ll x=;register bool y=;
    while(ch!='-' && (ch>'' || ch<''))ch=getchar();
    if(ch=='-')ch=getchar(),y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=getchar();
    return y?x:-x;
}
inline ll lb(ll y,ll x)
{
    ll l=,r=gg[x].size()-;
    while(l<=r){ll mid=(l+r)>>;if(gg[x][mid]>=y)r=mid-;if(gg[x][mid]<y)l=mid+;}
    return l;
}
inline ll ub(ll y,ll x)
{
    ll l=,r=gg[x].size()-;
    while(l<=r){ll mid=(l+r)>>;if(gg[x][mid]>y)r=mid-;if(gg[x][mid]<=y)l=mid+;}
    return l;
}
inline void wk1(){ll l=read(),r=read(),c=read();if(gg[c].size()==){printf("0\n");return;}printf("%d\n",ub(r,c)-lb(l,c));}
inline void wk2(){ll x=read();if(a[x]==a[x+])return;gg[a[x]][lb(x,a[x])]=x+;gg[a[x+]][lb(x+,a[x+])]=x;swap(a[x],a[x+]);}

int main()
{
    n=read();m=read();rp(i,,n)a[i]=read(),gg[a[i]].push_back(i);
    while(m--){ll op=read();op==?wk1():wk2();}
    return ;
}

放下手写二分写挂了的代码,,,我找不出错,,,,就很难过TT