Mayor's posters（线段树 + 离散化 + 区间更新）

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40494		Accepted: 11780

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 

• Every candidate can place exactly one poster on the wall. 
• All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 
• The wall is divided into segments and the width of each segment is one byte. 
• Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l_i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l_i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l_i, l_i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 

Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

 

      题意：

      给出 T，代表有 T 组数据。每组数据给出 n（1 ~ 10000）块板，每块板都有一个左右边界 （1 ~ 10 ^ 7），新放进去的板会覆盖掉原来的板，求出最后能看见的板能有多少种。

 

      思路：

      线段树。区间更新。维护线段的覆盖颜色，这里有个问题就是这里的覆盖的编号指的是线段的编号，而不是点的编号，比如样例

      3

      1 10

      1 4

      5 10    答案是2

      3

      1 10

      1 4

      6 10   答案是3

      这样的话，离散化之后就会带来问题，离散化后同样都是 1 2 3 4，得出来的答案永远都是2，故答案错误。

      所以当把左右边界放到一个数组之后，比如 1  4 6 10，将相邻两个数相差 1 的添加多一个中间值，处理后就是1,2,4,5,6,10，之后再做离散化求覆盖情况即可。

 

      AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 10005;

int cl[MAX], cr[MAX];
int num[MAX * 100], ans;

int color[MAX * 100];
int col_temp[MAX], sum;

void push_down(int node, int l, int r) {
        if (color[node]) {
                int mid = (r + l) >> 1;
                color[node << 1] = color[node];
                color[node << 1 | 1] = color[node];
                color[node] = 0;
        }
}

void build (int node, int l, int r) {
        if (r == l) {
                color[node] = 0;
        } else {
                int mid = (r + l) >> 1;
                build (node << 1, l, mid);
                build (node << 1 | 1, mid + 1, r);
                color[node] = 0;
        }
}

void update(int node, int l, int r, int al, int ar, int col) {
        if (ar < l || al > r) return;
        if (al <= l && ar >= r) {
                color[node] = col;
                return;
        }
        if (r == l) return;

        push_down(node, l, r);
        int mid = (r + l) >> 1;
        update(node << 1, l, mid, al, ar, col);
        update(node << 1 | 1, mid + 1, r, al, ar, col);
}

void query(int node, int l, int r) {
        if (color[node]) {
                if (!col_temp[ color[node] ]) {
                        ++sum;
                        col_temp[ color[node] ] = 1;
                }
                return;
        }
        if (r == l) return;

        int mid = (l + r) >> 1;
        query(node << 1, l, mid);
        query(node << 1 | 1, mid + 1, r);
}

int main() {
        int t;
        scanf("%d", &t);

        while (t--) {
                int n;
                scanf("%d", &n);

                ans = 0;
                for (int i = 0; i < n; ++i) {
                        scanf("%d%d", &cl[i], &cr[i]);
                        num[ans++] = cl[i];
                        num[ans++] = cr[i];
                }

                sort(num, num + ans);

                int a = ans;
                for (int i = 1; i < ans; ++i) {
                        if (num[i] - num[i - 1] > 1) {
                                num[a++] = num[i - 1] + 1;
                        }
                }
                ans = a;

                sort(num, num + ans);
                ans = unique(num, num + ans) - num;

                build(1, 1, ans);
                for (int i = 0; i < n; ++i) {
                        int l = lower_bound(num, num + ans, cl[i]) - num;
                        int r = lower_bound(num, num + ans, cr[i]) - num;
                        update(1, 1, ans, l + 1, r + 1, i + 1);
                }

                memset(col_temp, 0, sizeof(col_temp));
                sum = 0;
                query(1, 1, ans);

                printf("%d\n", sum);
        }

        return 0;
}