ZOJ-1610 Count the Colors (离散+线段树)
来源:互联网 发布:淘宝复制链接打不开啊 编辑:程序博客网 时间:2024/06/05 16:48
Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.
Your task is counting the segments of different colors you can see at last.
Input
The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.
Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:
x1 x2 c
x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.
All the numbers are in the range [0, 8000], and they are all integers.
Input may contain several data set, process to the end of file.
Output
Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.
If some color can't be seen, you shouldn't print it.
Print a blank line after every dataset.
Sample Input
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define maxn 8001int c[maxn << 3], cnt[maxn], l[maxn], r[maxn], v[maxn], ls[maxn << 2];void pushdown(int o, int l, int r){if(l == r || c[o] == -1) return;c[o << 1] = c[o << 1 | 1] = c[o];c[o] = -1;}void add(int o, int l, int r, int L, int R, int v){pushdown(o, l, r);if(l >= L && r <= R){c[o] = v;return;}int mid = l + r >> 1;if(mid >= L) add(o << 1, l, mid, L, R, v);if(mid < R) add(o << 1 | 1, mid + 1, r, L, R, v);}int query(int o, int l, int r, int id){if(l == r) return c[o];pushdown(o, l, r);int mid = l + r >> 1;if(id <= mid) return query(o << 1, l, mid, id);else return query(o << 1 | 1, mid + 1, r, id);}int main(){int n, x, y;while(scanf("%d", &n) != EOF){memset(c, -1, sizeof(c));memset(cnt, 0, sizeof(cnt));int tot = 0;for(int i = 1; i <= n; ++i){scanf("%d %d %d", &l[i], &r[i], &v[i]);ls[++tot] = ++l[i];ls[++tot] = r[i];}sort(ls + 1, ls + 1 + tot);int num = unique(ls + 1, ls + 1 + tot) - ls;int m = num;for(int i = 2; i < num; ++i){if(ls[i] - ls[i - 1] > 1){ls[m++] = ls[i - 1] + 1;}}sort(ls + 1, ls + m);for(int i = 1; i <= n; ++i){x = lower_bound(ls + 1, ls + m, l[i]) - ls;y = lower_bound(ls + 1, ls + m, r[i]) - ls;add(1, 1, m, x, y, v[i]);}x = query(1, 1, m, 1);for(int i = 2; i <= m; ++i){y = query(1, 1, m, i);if(x != y){if(x != -1) cnt[x]++;x = y;}}if(x != -1) cnt[x]++;for(int i = 0; i <= 8000; ++i){if(cnt[i] != 0){printf("%d %d\n", i, cnt[i]);}}puts("");}}/*题意: 按顺序对不同区间刷不同颜色,问最后整个区间有多少个颜色段。 思路: 线段树区间修改区间查询。需要考虑的是区间长度未知,需要离散化一下,离散的时候一定注意 如果差值大于2的情况不要让它们离散后连续了,比如[1,2],[4,5]很可能离散成[1,2],[3,4],这样会导致答案错误。 */
- ZOJ-1610 Count the Colors (离散+线段树)
- ZOJ 1610 Count the Colors(线段覆盖着色:离散化)
- ZOJ 1610Count the Colors//线段树
- zoj 1610 Count the Colors【线段树】
- zoj 1610 Count the Colors[线段树]
- 【线段树】Count the Colors (zoj 1610)
- zoj 1610 Count the Colors(线段树)
- zoj 1610 Count the Colors 线段树
- ZOJ - 1610 Count the Colors(线段树)
- ZOJ 1610 Count the Colors【线段树】
- ZOJ----1610-Count the Colors(线段树)
- zoj 1610 Count the Colors(线段树)
- ZOJ 1610 Count the Colors【线段树】
- ZOJ 1610 Count the Colors 线段树 染色问题
- zoj 1610 Count the Colors(线段树延迟更新)
- ZOJ 题目1610 Count the Colors(线段树)
- zoj - 1610 - Count the Colors(线段树)
- ZOJ 1610Count the Colors 线段树_区间更新
- 欢迎使用CSDN-markdown编辑器
- 160
- Two Rabbits HDU
- Spring使用Quartz定时调度Job无法Autowired注入Service的解决方案
- VMware中解决ubuntu不能连接网络问题---NAT模式
- ZOJ-1610 Count the Colors (离散+线段树)
- 选择性搜索(selective search)
- 安卓集成 极光推送 自动集成 过程
- Android 开发问题 —— 给 CollapsingToolbarLayout 设置 addOnOffsetChangedListener
- 接口
- Spring+Quartz定时器结合调用service
- C语言__指针
- Java和C++的区别!
- 线性代数(二)