HDU 5023 A Corrupt Mayor's Performance Art （線段樹）
A Corrupt Mayor's Performance Art
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 255 Accepted Submission(s): 114
Problem Description
Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money.
Because a lot of people praised mayor X's painting(of course, X was a mayor), mayor X believed more and more that he was a very talented painter. Soon mayor X was not satisfied with only making money. He wanted to be a famous painter. So he joined the local
painting associates. Other painters had to elect him as the chairman of the associates. Then his painting sold at better price.
The local middle school from which mayor X graduated, wanted to beat mayor X's horse fart(In Chinese English, beating one's horse fart means flattering one hard). They built a wall, and invited mayor X to paint on it. Mayor X was very happy. But he really had
no idea about what to paint because he could only paint very abstract paintings which nobody really understand. Mayor X's secretary suggested that he could make this thing not only a painting, but also a performance art work.
This was the secretary's idea:
The wall was divided into N segments and the width of each segment was one cun(cun is a Chinese length unit). All segments were numbered from 1 to N, from left to right. There were 30 kinds of colors mayor X could use to paint the wall. They named those colors
as color 1, color 2 .... color 30. The wall's original color was color 2. Every time mayor X would paint some consecutive segments with a certain kind of color, and he did this for many times. Trying to make his performance art fancy, mayor X declared that
at any moment, if someone asked how many kind of colors were there on any consecutive segments, he could give the number immediately without counting.
But mayor X didn't know how to give the right answer. Your friend, Mr. W was an secret officer of anti-corruption bureau, he helped mayor X on this problem and gained his trust. Do you know how Mr. Q did this?
Input
There are several test cases.
For each test case:
The first line contains two integers, N and M ,meaning that the wall is divided into N segments and there are M operations(0 < N <= 1,000,000; 0
Then M lines follow, each representing an operation. There are two kinds of operations, as described below:
1) P a b c
a, b and c are integers. This operation means that mayor X painted all segments from segment a to segment b with color c ( 0 < a<=b <= N, 0 < c <= 30).
2) Q a b
a and b are integers. This is a query operation. It means that someone asked that how many kinds of colors were there from segment a to segment b ( 0 < a<=b <= N).
Please note that the operations are given in time sequence.
The input ends with M = 0 and N = 0.
Output
For each query operation, print all kinds of color on the queried segments. For color 1, print 1, for color 2, print 2 ... etc. And this color sequence must be in ascending order.
Sample Input
5 10
P 1 2 3
P 2 3 4
Q 2 3
Q 1 3
P 3 5 4
P 1 2 7
Q 1 3
Q 3 4
P 5 5 8
Q 1 5
0 0
Sample Output
4
3 4
4 7
4
4 7 8
題意：
在片段上著色，有兩種操作，如下：
第一種：P a b c 把 a 片段至 b 片段的顏色都變為 c 。
第二種：Q a b 詢問 a 片段至 b 片段有哪些顏色，把這些顏色按從小到大的編號輸入，不要有重復
片段上默認的初始顏色為編號2的顏色。
思路：
1、利用線段樹進行區間覆蓋。
2、在每個節點上附上一個60大小的數組，便於將左右兒子的顏色編號合並到自己節點的數組上（因為顏色編號最多為30）。
3、每次合並左右兒子的顏色編號都需要一次排序和去重，防止記錄數超過數組大小。
4、由於0
/*************************************************************************
> File Name: hdu5023.cpp
> Author: Bslin
> Mail: [email protected]
> Created Time: 2014年09月20日 星期六 21時24分19秒
************************************************************************/
#include
#include
#include
using namespace std;
const int N = 200010;
const int M = 100010;
struct node {
int l, r;
int cnt;
bool flag;
int num[70];
} tree[N << 2];
int ans[70], tot;
int Hash[N], a[M], b[M], c[M];
char op[M];
void pushup(int p) {
int i, cnt;
memset(tree[p].num, 0, sizeof(tree[p].num));
for(i = 0; i < tree[p << 1].cnt; i ++) {
tree[p].num[i] = tree[p << 1].num[i];
}
int len = tree[p << 1].cnt;
for(i = 0; i < tree[p << 1 | 1].cnt; i ++) {
tree[p].num[len + i] = tree[p << 1 | 1].num[i];
}
cnt = tree[p << 1].cnt + tree[p << 1 | 1].cnt;
sort(tree[p].num, tree[p].num + cnt);
tree[p].cnt = unique(tree[p].num, tree[p].num + cnt) - tree[p].num;
}
void pushdown(int p) {
if(tree[p].flag) {
tree[p << 1].flag = tree[p << 1 | 1].flag = 1;
tree[p].flag = 0;
memset(tree[p << 1].num, 0, sizeof(tree[p << 1].num));
memset(tree[p << 1 | 1].num, 0, sizeof(tree[p << 1 | 1].num));
tree[p << 1].cnt = 1;
tree[p << 1 | 1].cnt = 1;
tree[p << 1].num[0] = tree[p].num[0];
tree[p << 1 | 1].num[0] = tree[p].num[0];
}
}
void build(int l, int r, int p) {
tree[p].l = l;
tree[p].r = r;
tree[p].flag = 0;
tree[p].cnt = 1;
memset(tree[p].num, 0, sizeof(tree[p].num));
tree[p].num[0] = 2;
if(l == r) {
return ;
}
int mid = (l + r) >> 1;
build(l, mid, p << 1);
build(mid + 1, r, p << 1 | 1);
}
void update(int l, int r, int val, int p) {
if(tree[p].l == l && tree[p].r == r) {
memset(tree[p].num, 0, sizeof(tree[p].num));
tree[p].cnt = 1;
tree[p].num[0] = val;
tree[p].flag = 1;
return ;
}
pushdown(p);
int mid = (tree[p].l + tree[p].r) >> 1;
if(r <= mid) update(l, r, val, p << 1);
else if(l > mid) update(l,r,val,p << 1 | 1);
else {
update(l, mid, val, p << 1);
update(mid + 1, r, val, p << 1 | 1);
}
pushup(p);
}
void query(int l, int r, int p) {
if(tree[p].l == l && tree[p].r == r) {
int i;
for(i = 0; i < tree[p].cnt; i ++) {
ans[tot++] = tree[p].num[i];
}
sort(ans, ans + tot);
tot = unique(ans, ans + tot) - ans;
return ;
}
pushdown(p);
int mid = (tree[p].l + tree[p].r) >> 1;
if(mid >= r) query(l, r, p << 1);
else if(l > mid) query(l, r, p << 1 | 1);
else {
query(l, mid, p << 1);
query(mid + 1, r, p << 1 | 1);
}
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in", "r", stdin);
#endif
int n, m, i, j;
int hh;
while(scanf("%d%d", &n, &m) != EOF) {
if(n == 0 && m == 0) break;
hh = 0;
getchar();
for(i = 0; i < m; i ++) {
scanf("%c", &op[i]);
if(op[i] == 'P') {
scanf("%d%d%d", &a[i], &b[i], &c[i]);
Hash[hh++] = a[i];
Hash[hh++] = b[i];
} else {
scanf("%d%d", &a[i], &b[i]);
Hash[hh++] = a[i];
Hash[hh++] = b[i];
}
getchar();
}
sort(Hash, Hash + hh);
hh = unique(Hash, Hash + hh) - Hash;
build(1, hh, 1);
for(i = 0; i < m; i ++) {
a[i] = lower_bound(Hash, Hash + hh, a[i] ) - Hash + 1;
b[i] = lower_bound(Hash, Hash + hh, b[i] ) - Hash + 1;
}
for(i = 0; i < m; i++) {
if(op[i] == 'P') {
update(a[i], b[i], c[i], 1);
} else {
memset(ans, 0, sizeof(ans));
tot = 0;
query(a[i], b[i], 1);
for(j = 0; j < tot; j ++) {
if(j == 0) printf("%d", ans[0]);
else printf(" %d", ans[j]);
}
printf("\n");
}
}
}
}
