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

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题意：一长由N小段组成的长条，每小段的初始颜色为2。现执行M个操作，每个操作是以下两种中的一种(0 < N <= 1,000,000; 0<M<=100,000) ：

P a b c ——> 将段a到段b涂成颜色c，c是1, 2, ... 30中的一种(0 < a<=b <= N, 0 < c <= 30)。

Q a b ——> 问段a到段b之间有哪几种颜色，按颜色大小从小到大输出(0 < a<=b <= N, 0 < c <= 30)。

——>>很明显此题可以用线段树实现Mlog(N)的解法。。（看完输入就敲题了，把初始颜色设为了无，耗了好多时间：看题要仔细啊。。）

#include <cstdio>#include <set>#define lc (o << 1)#define rc ((o << 1) | 1)using std::set;const int MAXN = 1000000 + 10;int g_nColor[MAXN << 2];set<int> g_sRet;void Initialize(){g_sRet.clear();}void Build(int o, int L, int R){g_nColor[o] = 2;if (L == R){return;}int M = (L + R) >> 1;Build(lc, L, M);Build(rc, M + 1, R);}void Pushdown(int o){if (g_nColor[o]){g_nColor[lc] = g_nColor[rc] = g_nColor[o];g_nColor[o] = 0;}}void Update(int o, int L, int R, int ql, int qr, int nColor){if (ql <= L && R <= qr){g_nColor[o] = nColor;return;}Pushdown(o);int M = (L + R) >> 1;if (ql <= M){Update(lc, L, M, ql, qr, nColor);}if (qr > M){Update(rc, M + 1, R, ql, qr, nColor);}}void Query(int o, int L, int R, int ql, int qr){if (ql <= L && R <= qr && g_nColor[o]){g_sRet.insert(g_nColor[o]);return;}if (L == R){return;}Pushdown(o);int M = (L + R) >> 1;if (ql <= M){Query(lc, L, M, ql, qr);}if (qr > M){Query(rc, M + 1, R, ql, qr);}}void Output(){if (g_sRet.empty()){return;}set<int>::const_iterator iter = g_sRet.begin();size_t nCnt = 0;for (; nCnt < g_sRet.size() - 1; ++nCnt, ++iter){printf("%d ", *iter);}printf("%d\n", *iter);}int main(){int N, M, a, b, c;char chOperation;while (scanf("%d%d", &N, &M) == 2 && (N && M)){Build(1, 1, N);while (M--){getchar();chOperation = getchar();if (chOperation == 'P'){scanf("%d%d%d", &a, &b, &c);Update(1, 1, N, a, b, c);}else{scanf("%d%d", &a, &b);Initialize();Query(1, 1, N, a, b);Output();}}}return 0;}

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