[POJ2777]Count Color（线段树）

题目链接：http://poj.org/problem?id=2777

给你一个长为L想线段，向上面染色，颜色不超过３０种，一共有Ｏ次操作，操作有两种：

C a b c　在[a,b]上染上c颜色

P a b　查询[a,b]上所有颜色数。

思路：线段树维护每个线段上颜色种类，用位来存颜色。好题。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &(a))
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long Uint;
 typedef pair<int, int> pii;
 typedef pair<LL, LL> pLL;
 typedef pair<string, LL> psi;
 typedef map<string, LL> msi;
 typedef vector<LL> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 const int maxn = ;
 int sum[maxn<<];
 int add[maxn<<];
 int L, T, O;

 void pushUP(int rt) {
     sum[rt] = sum[lrt] | sum[rrt];
 }

 void pushDOWN(int rt) {
     if(add[rt] != -) {
         sum[lrt] = sum[rrt] = ( << add[rt]);
         add[lrt] = add[rrt] = add[rt];
         add[rt] = -;
     }
 }

 void build(int l, int r, int rt) {
     sum[rt] = ; add[rt] = ;
     if(l == r) return;
     int m = (l + r) >> ;
     build(l, m, lrt);
     build(m+, r, rrt);
 }

 void update(int L, int R, int c, int l, int r, int rt) {
     if(L <= l && r <= R) {
         add[rt] = c;
         sum[rt] =  << c;
         return;
     }
     pushDOWN(rt);
     int m = (l + r) >> ;
     if(L <= m) update(L, R, c, lson);
     if(m < R) update(L, R, c, rson);
     pushUP(rt);
 }

 int query(int L, int R, int l, int r, int rt) {
     if(L <= l && r <= R) return sum[rt];
     pushDOWN(rt);
     int m = (l + r) >> ;
     int ret = ;
     if(L <= m) ret |= query(L, R, lson);
     if(m < R) ret |= query(L, R, rson);
     return ret;
 }

 int main() {
 //    FRead();
     char cmd[];
     int l, r, c;
     while(~scanf("%d%d%d", &L, &T, &O)) {
         build(, L, );
         W(O) {
             Rs(cmd);
             if(cmd[] == 'C') {
                 scanf("%d%d%d",&l,&r,&c);
                 if(l > r) swap(l, r);
                 update(l, r, c, , L, );
             }
             else {
                 scanf("%d%d",&l,&r);
                 if(l > r) swap(l, r);
                 int p = query(l, r, , L, );
                 int ret = ;
                 while(p) {
                     if(p & ) ret++;
                     p >>= ;
                 }
                 printf("%d\n", ret);
             }
         }
     }
     RT ;
 }