zoj 1610 Count the Colors （线段树区间覆盖）

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

<b< dd="">

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

题意：

在长度8000的一条线段上染色，后面染上的颜色会把先前的覆盖，求染完色之后，每颜色分别存在几段。

在输入区间长度的时候，令左端点加一，这样就可以把区间看成是不连续的点。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int mx = 8100;int node[4 * mx],num[mx],last;void pushdown(int rt){if(node[rt] != -1){node[2 * rt] = node[2 * rt + 1] = node [rt];node[rt] = -1;return ;}}void update(int st, int en, int p, int l, int r, int rt){if(st <= l && r <= en){node[rt] = p;return;}pushdown(rt);int m = (l + r) / 2;if(st <= m) update(st, en, p, l, m, 2 * rt);if(m < en) update(st, en, p, m + 1, r, 2 * rt + 1);}void query(int l, int r, int rt){if(l == r){ if(node[rt] != -1 && node[rt] != last)num[node[rt]]++;last = node[rt];return ;} pushdown(rt);int m =(l + r) / 2;query(l, m, 2 * rt);query(m+1, r, 2 * rt + 1);}int main(){int n;while(scanf("%d", &n) != EOF){memset(num, 0, sizeof(num));memset(node, -1, sizeof(node)); last = -1;int a, b, c;while(n--){scanf("%d%d%d", &a, &b, &c);if(a < b) update(a + 1, b, c, 1, 8000, 1);}query(1, 8000, 1);for(int i = 0; i < mx; i++){if(num[i]) printf("%d %d\n", i, num[i]);}puts("");}return 0;}