BZOJ 2120: 数颜色

2120: 数颜色

Time Limit: 6 Sec  Memory Limit: 259 MB
Submit: 3623  Solved: 1396
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Description

墨墨购买了一套N支彩色画笔（其中有些颜色可能相同），摆成一排，你需要回答墨墨的提问。墨墨会像你发布如下指令： 1、 Q L R代表询问你从第L支画笔到第R支画笔中共有几种不同颜色的画笔。 2、 R P Col 把第P支画笔替换为颜色Col。为了满足墨墨的要求，你知道你需要干什么了吗？

Input

第1行两个整数N，M，分别代表初始画笔的数量以及墨墨会做的事情的个数。第2行N个整数，分别代表初始画笔排中第i支画笔的颜色。第3行到第2+M行，每行分别代表墨墨会做的一件事情，格式见题干部分。

Output

对于每一个Query的询问，你需要在对应的行中给出一个数字，代表第L支画笔到第R支画笔中共有几种不同颜色的画笔。

Sample Input

6 5
1 2 3 4 5 5
Q 1 4
Q 2 6
R 1 2
Q 1 4
Q 2 6

Sample Output

4
4
3
4

HINT

对于100%的数据，N≤10000，M≤10000，修改操作不多于1000次，所有的输入数据中出现的所有整数均大于等于1且不超过10^6。

2016.3.2新加数据两组by Nano_Ape

Source

 

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这真是一道好题，貌似有好多种不同的做法。有人用树套树过的，有人用整体二分过的，有人用带修莫队过的，还有像我一样的蒟蒻用常数优化暴力过的。

虽然2016有新加数据，然而依然卡不住暴力，23333。

RunID	User	Problem	Result	Memory	Time	Language	Code_Length	Submit_Time
1734100	YOUSIKI	2120	Accepted	34496 kb	2340 ms	C++/Edit	1411 B	2016-12-11 12:45:43

 #include <bits/stdc++.h>

 const int siz = ;

 char buf[siz], *bit = buf;

 inline int nextInt (void) {
     register int ret = ;
     register int neg = ;

     while (*bit < '')
         if (*bit++ == '-')
             neg ^= true;

     while (*bit >= '')
         ret = ret* + *bit++ - '';

     return neg ? -ret : ret;
 }

 inline char nextChar (void) {
     while (*bit != 'Q' && *bit != 'R')++bit;
     return *bit ++;
 }

 const int maxn =  + ;

 int n, m;
 int cases;
 int total;
 int answer;
 int col[maxn];
 int map[maxn];
 int vis[maxn];

 signed main (void) {
     fread (buf, , siz, stdin);

     n = nextInt ();
     m = nextInt ();

     for (register int i = ; i <= n; ++i) {
         int color = nextInt ();
         if (map[color] == )
             map[color] = ++total;
         col[i] = map[color];
     }

     for (register int i = ; i <= n; ++i) {
         switch (nextChar ()) {
             case 'R' : {
                 int a = nextInt ();
                 int b = nextInt ();
                 if (map[b] == )
                     map[b] = ++total;
                 col[a] = map[b];
                 break;
             }
             case 'Q' : {
                 int a = nextInt ();
                 int b = nextInt ();
                 answer = , ++cases;
                 for (register int j = a; j <= b; ++j)
                     if (vis[col[j]] != cases)++answer, vis[col[j]] = cases;
                 printf("%d\n", answer);
             }
         }
     }
 }

然而不太理解为啥离散化一下就有这么大的用处。

RunID	User	Problem	Result	Memory	Time	Language	Code_Length	Submit_Time
1734122	YOUSIKI	2120	Time_Limit_Exceed	26684 kb	7632 ms	C++/Edit	1289 B	2016-12-11 13:00:40

 #include <bits/stdc++.h>

 const int siz = ;

 char buf[siz], *bit = buf;

 inline int nextInt (void) {
     register int ret = ;
     register int neg = ;

     while (*bit < '')
         if (*bit++ == '-')
             neg ^= true;

     while (*bit >= '')
         ret = ret* + *bit++ - '';

     return neg ? -ret : ret;
 }

 inline char nextChar (void) {
     while (*bit != 'Q' && *bit != 'R')++bit;
     return *bit ++;
 }

 const int maxn =  + ;

 int n, m;
 int cases;
 int answer;
 int col[maxn];
 int vis[maxn];

 signed main (void) {
     fread (buf, , siz, stdin);

     n = nextInt ();
     m = nextInt ();

     for (register int i = ; i <= n; ++i)
         col[i] = nextInt ();

     for (register int i = ; i <= m; ++i) {
         switch (nextChar ()) {
             case 'R' : {
                 int a = nextInt ();
                 int b = nextInt ();
                 col[a] = b;
                 break;
             }
             case 'Q' : {
                 int a = nextInt ();
                 int b = nextInt ();
                 answer = , ++cases;
                 for (register int j = a; j <= b; ++j)
                     if (vis[col[j]] != cases)++answer, vis[col[j]] = cases;
                 printf ("%d\n", answer);
             }
         }
     }
 }

当然，小生来写这道题不是为了练习暴力的，是为了学习带修莫队的。

有别于普通的莫队，带修莫队多了一维对时间的要求，每个询问可以表示为[l,r,t]，表示区间需要我们维护到区间[l,r]在时刻t的存在情况。方法是，对于l,r还按照普通莫队的对l分组方式分组，组内对r排序，转移的时候暴力调整时间即可。貌似为了达到最优秀的复杂度，分组时的大小需要调整一下，但这题显然没有那么苛刻啦，o(*￣▽￣*)ブ

 #include <bits/stdc++.h>

 const int siz = ;

 char buf[siz], *bit = buf;

 inline int nextInt (void) {
     register int ret = ;
     register int neg = ;

     while (*bit < '')
         if (*bit++ == '-')
             neg ^= true;

     while (*bit >= '')
         ret = ret* + *bit++ - '';

     return neg ? -ret : ret;
 }

 inline char nextChar (void) {
     while (*bit != 'Q' && *bit != 'R')++bit;
     return *bit ++;
 }

 const int maxn =  + ;
 const int maxm =  + ;

 int n, m;
 int l, r;
 int s, t;
 int answer;
 int col[maxn];
 int vis[maxm];

 struct query {
     int l, r, id, ans;
 }qry[maxn]; int q;

 struct change {
     int p, a, b, id;
 }chg[maxn]; int c;

 inline bool cmp_lr (const query &A, const query &B) {
     if (A.l / s != B.l / s)
         return A.l < B.l;
     else
         return A.r < B.r;
 }

 inline bool cmp_id (const query &A, const query &B) {
     return A.id < B.id;
 }

 inline void removeColor (int c) {
     if (--vis[c] == )--answer;
 }

 inline void insertColor (int c) {
     if (++vis[c] == )++answer;
 }

 inline void removeChange (change &c) {
     col[c.p] = c.a;
     if (c.p >= l && c.p <= r) {
         removeColor(c.b);
         insertColor(c.a);
     }
 }

 inline void insertChange (change &c) {
     col[c.p] = c.b;
     if (c.p >= l && c.p <= r) {
         removeColor(c.a);
         insertColor(c.b);
     }
 }

 inline void solve (query &q) {
     while (t >  && chg[t].id > q.id)
         removeChange (chg[t--]);
     while (t < c && chg[t + ].id < q.id)
         insertChange (chg[++t]);

     while (l < q.l)removeColor (col[l++]);
     while (l > q.l)insertColor (col[--l]);
     while (r < q.r)insertColor (col[++r]);
     while (r > q.r)removeColor (col[r--]);

     q.ans = answer;
 }

 signed main (void) {
     fread (buf, , siz, stdin);

     n = nextInt ();
     m = nextInt ();

     for (register int i = ; i <= n; ++i)
         col[i] = nextInt ();

     for (register int i = ; i <= m; ++i) {
         switch (nextChar ()) {
             case 'R' : {
                 chg[++c].id = i;
                 chg[c].p = nextInt ();
                 chg[c].b = nextInt ();
                 chg[c].a = col[chg[c].p];
                 col[chg[c].p] = chg[c].b;
                 break;
             }
             case 'Q' : {
                 qry[++q].id = i;
                 qry[q].l = nextInt ();
                 qry[q].r = nextInt ();
             }
         }
     }

     s = sqrt (n); t = c; answer = ; l = ; r = ;

     std::sort (qry + , qry +  + q, cmp_lr);

     for (register int i = ; i <= q; ++i)solve(qry[i]);

     std::sort (qry + , qry +  + q, cmp_id);

     for (register int i = ; i <= q; ++i)
         printf ("%d\n", qry[i].ans);
 }

@Author: YouSiki