HDU 5023 A Corrupt Mayor's Performance Art

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题意：

对一个栅栏有两种操作，一种是把从a到b的所有栅栏都刷成c，另一种是询问从a到b有多少种颜色。（栅栏初始的颜色是2）。

思路：

线段树的一道题。

利用位运算存储颜色，反正颜色只用30种，int正好够用。

利用延迟标记更新节点。

想下延伸的时候，需要把本身就置成要修改的颜色，因为标记的时候，这个节点下面的节点就应该都被修改。

Code：

#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>#include<string>#include<queue>#include<stack>#include<bitset>#include<set>#include<map>#include<cctype>#include<vector>//#define TEST#define LL long long#define Mt(f, x) memset(f, x, sizeof(f));#define rep(i, s, e) for(int i = (s); i <= (e); ++i)#ifdef TEST    #define See(a) cout << #a << " = " << a << endl;    #define See2(a, b) cout << #a << " = " << a << ' ' << #b << " = " << b << endl;    #define debug(a, s, e){ rep(_i, s, e) {cout << a[_i] << ' '; }cout << endl;}    #define debug2(a, s, e, ss, ee) rep(i_, s, e) {debug(a[i_], ss, ee);}#else    #define See(a)    #define See2(a, b)    #define debug(a, s, e)    #define debug2(a, s, e, ss, ee)#endifconst int MAX = 2e9;const int MIN = -2e9;const int PI = acos(-1.0);const double eps = 1e-8;using namespace std;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const int N = 1000000 + 5;int col[N << 2], dif[N << 2];void pushUp(int rt){//    See(rt);    col[rt] = col[rt << 1] | col[rt << 1 | 1];}void build(int l, int r, int rt){    if(l == r)    {        col[rt] = 1 << 2;        return ;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    pushUp(rt);}void pushDown(int rt){//    See(rt);    if(dif[rt])    {        dif[rt << 1] = dif[rt << 1 | 1] = dif[rt];        col[rt << 1] = col[rt << 1 | 1] = dif[rt];        dif[rt] = 0;    }}void update(int L, int R, int c, int l, int r, int rt){    if(l >= L && r <= R)    {//        See(rt);        dif[rt] = 1 << c;        col[rt] = dif[rt];        return ;    }    pushDown(rt);    int m = (l + r) >> 1;    if(L <= m) update(L, R, c, lson);    if(R > m) 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 col[rt];    }    pushDown(rt);    int ret = 0;    int m = (l + r) >> 1;    if(L <= m) ret |= query(L, R, lson);    if(R > m) ret |= query(L, R, rson);    return ret;}void print(int t){    bool first = true;    for(int i = 1; i <= 30; ++i)    {        if(t & (1 << i))        {            printf(first ? "%d" : " %d", i);            first = false;        }    }    printf("\n");}int main(){    int n, q;    while(~scanf("%d%d", &n, &q))    {        Mt(dif, 0);        if(n + q == 0)        {            break;        }        build(1, n, 1);        while(q--)        {            char op[2];            int a, b, c;            scanf("%s%d%d", op, &a, &b);            if(op[0] == 'P')            {                scanf("%d", &c);                update(a, b, c, 1, n, 1);            }            else            {                int t = query(a, b, 1, n, 1);                print(t);            }        }    }    return 0;}

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