POJ 2528 线段树+离散化
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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.
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.![技术图片]()
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
题目大意：给你一个长板子，然后有n个人在区间[li, ri]范围内贴海报，问贴完之后还能看见多少张海报。
思路分析：线段树区间更新问题，但是要注意，给的长度的可能非常大，有1e9，但是我们注意到一共最多只有2e4个点，因此我们可以用离散化的思想先对区间进行处理
这里给两个例子，普通的离散化会错误。
eg1:1-10 1-4 5-10
eg2:1-10 1-4 6-10
普通离散化后都变成了[1,4][1,2][3,4]
例子1被完全覆盖了，而例子2没有被完全覆盖
但在压缩时，我们将仍然漏在外面的5给压缩掉了，所以我们在所有相隔距离大于1的点之间都插入一个单点（让它充当那段被裸露区间），然后此问题得以解决。
具体可去 https://blog.csdn.net/qq_35802619/article/details/98326267 了解
另外对c++中的unique函数 https://www.cnblogs.com/hua-dong/p/7943983.html
下面给出代码
#include <iostream> #include <algorithm> #include <string.h> #include <cstdio> #include <string> #include <cmath> #include <vector> #include <stack> #include <queue> #include <stack> #include <list> #include <map> #include <set> //#include <unordered_map> #define Fbo friend bool operator < (node a, node b) #define mem(a, b) memset(a, b, sizeof(a)) #define FOR(a, b, c) for(int a = b; a <= c; a++) #define RFOR(a,b, c) for(int a = b; a >= c; a--) #define off ios::sync_with_stdio(0) bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; } using namespace std; typedef pair<int, int> pii; typedef long long ll; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; const int Maxn = 2e5 + 5; const double pi = acos(-1.0); const double eps = 1e-8; ll a[Maxn]; ll li[Maxn], ri[Maxn]; vector<ll>num; struct node { ll l, r;//区间[l,r] ll add;//区间的延时标记 ll sum;//区间和 }tree[Maxn * 4 + 9]; ll n; void pushup(ll rt) { //向上更新 tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; } void pushdown(ll rt) {//向下更新 if (tree[rt].add) { //当我们的标记不是0，说明该段染色，所以往下传递 tree[rt << 1].add = tree[rt].add; tree[rt << 1 | 1].add = tree[rt].add; //这里不能用加 tree[rt].add = 0; //取消本层标记 } } void BuildTree(ll l, ll r, ll rt) { tree[rt].l = l; tree[rt].r = r; tree[rt].add = 0; if (l == r) { //tree[rt] = {l,r,0}; return; } ll mid = (l + r) >> 1; BuildTree(l, mid, rt << 1); //递归左子树 BuildTree(mid + 1, r, (rt << 1) + 1); //递归右子树 pushup(rt);//向上更新 } void updata(ll l, ll r, ll val, ll rt) { //root为结点 区间更新 //cout << l << " " << tree[rt].l << "----" << r << " " << tree[rt].r << endl; if (l <= tree[rt].l && r >= tree[rt].r) { tree[rt].add = val;//延时标记 return; } pushdown(rt); ll mid = (tree[rt].l + tree[rt].r) >> 1; if (l <= mid) { updata(l, r, val, rt << 1); } if (r > mid) { updata(l, r, val, rt << 1 | 1); } pushup(rt);//向上更新 可加可不加 本题并不需要根据儿子节点来更新父亲节点也能做 } void query(ll l, ll r, ll rt) { ////这个询问和普通线段树的询问略有不同 if (tree[rt].add) { // 如果标记存在，说明这个区间是单色的 a[tree[rt].add] = 1; //将有被染色的颜色记录下来 return; } if (tree[rt].l == tree[rt].r) //一定要这个判断条件，否则你没有递归终止条件，就RE了 return; //这与普通线段树的格式有区别，但是原理相同 pushdown(rt); ll mid = (tree[rt].l + tree[rt].r) >> 1; if (l <= mid) { query(l, r, rt << 1);//如果是0，有两种情况，一个是没有染色，一个是有多种颜色 } if (r > mid) { query(l, r, rt << 1 | 1); } } ll solve() { //离散化操作和插点操作 ll x, y; sort(num.begin(), num.end()); //先排序 num.erase(unique(num.begin(), num.end()), num.end()); ////删除重复的元素 int m = num.size(); for (int i = 0; i < m - 1; i++) { if (num[i + 1] - num[i] > 1) num.push_back((num[i + 1] - 1));//所有相隔距离大于1的点之间都插入一个单点 } //（让它充当那段被裸露区间）,这样就不会漏掉点 sort(num.begin(), num.end()); BuildTree(0, num.size() - 1, 1); for (ll i = 1; i <= n; i++) { // 全开ll方便点 x = lower_bound(num.begin(), num.end(), li[i]) - num.begin(); y = lower_bound(num.begin(), num.end(), ri[i]) - num.begin(); updata(x, y, i, 1); } ll ans = 0; mem(a, 0); query(0, num.size() - 1, 1); FOR(i, 1, n) { if (a[i]) ans++; } return ans; } int main() { ll t; scanf("%lld", &t); while (t--) { scanf("%lld", &n); num.clear(); FOR(i, 1, n) { scanf("%lld%lld", &li[i], &ri[i]); num.push_back(li[i]); num.push_back(ri[i]); } printf("%lld ", solve()); } return 0; }
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