hdu 5023 A Corrupt Mayor's Performance Art(线段树)

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题目链接：hdu 5023 A Corrupt Mayor's Performance Art

题目大意：给定一段区间，初始颜色为2，接诊有N次操作，P：将区间l，r涂成c，Q：查询l，r之间有哪些颜色按字典序输出。

解题思路：因为颜色的种数只有30，对于每个节点之间用一个二进制数记录即可，水水的线段树。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 1000005;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)+1)struct Node {    int l, r, s, v;    void set (int l, int r, int s, int v) {        this->l = l;        this->r = r;        this->s = s;        this->v = v;    }    void set (int x) {        this->v = x;        this->s = x;    }}nd[maxn * 4];int N, M;vector<int> ans;void pushup (int u) {    nd[u].s = nd[lson(u)].s | nd[rson(u)].s;}void pushdown(int u) {    if (nd[u].v) {        nd[lson(u)].set(nd[u].v);        nd[rson(u)].set(nd[u].v);        nd[u].v = 0;    }}void build (int u, int l, int r) {    nd[u].set(l, r, 0, 0);    if (l == r) {        nd[u].s = 2;        return ;    }    int mid = (l + r) / 2;    build (lson(u), l, mid);    build (rson(u), mid + 1, r);    pushup(u);}void modify (int u, int l, int r, int x) {    if (l <= nd[u].l && nd[u].r <= r) {        nd[u].set(x);        return;;    }    pushdown(u);    int mid = (nd[u].l + nd[u].r) / 2;    if (l <= mid)        modify(lson(u), l, r, x);    if (r > mid)        modify(rson(u), l, r, x);    pushup(u);}int query(int u, int l, int r) {    if (l <= nd[u].l && nd[u].r <= r)        return nd[u].s;    pushdown(u);    int mid = (nd[u].l + nd[u].r) / 2, ret = 0;    if (l <= mid)        ret |= query(lson(u), l, r);    if (r > mid)        ret |= query(rson(u), l, r);    pushup(u);    return ret;}int main () {    while (scanf("%d%d", &N, &M) == 2 && N + M) {        build (1, 1, N);        int l, r, x;        char order[10];        for (int i = 0; i < M; i++) {            scanf("%s%d%d", order, &l, &r);            if (order[0] == 'P') {                scanf("%d", &x);                modify(1, l, r, 1<<(x-1));            } else {                ans.clear();                int s = query(1, l, r);                for (int i = 0; i < 30; i++)                    if (s & (1<<i))                        ans.push_back(i+1);                printf("%d", ans[0]);                for (int i = 1; i < ans.size(); i++)                    printf(" %d", ans[i]);                printf("\n");            }        }    }    return 0;}

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