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:
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.
- 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.
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：
大体是说有人要竞选需要投票什么的，不用理会，总之后面就是贴海报， 后面的能把前面的海报覆盖，海报每次贴不会到方格中间，
如果一个海报不是完全被贴上就算可以看到，问最后你能看到几张海报（英语渣，反正我是这么理解的）
输入
1 //T个样例 5 //下面5（n）张海报及其范围 1 <= n <= 10000 1 4 //海报的左右边界 1--10000000 2 6 8 10 3 4 7 10
输出
4 能看到4张海报
解题思路：
由于海报粘贴的方式，在外面的绝对不会被里面的遮住，所以倒着来覆盖海报比较方便
主要还是由于海报边界太大需要离散化，如果把海报的每个边界看成一个节点话最多需要的范围就成了20000 节点是80000
不过我不会离散化第一次看了课件，模仿着写的，不过听学长说貌似不算很好的离散化 或者完全不合格的离散化
我用的这种离散化还是要开一个100000000的一维数组 虽然开的下，这道题也能解决，不过学长说如果范围是几亿的话就没办法了
不过这道题还是能过的，新的离散化方法明天再搞吧
下面是代码
/*毕竟我是王宇浩*/ #include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <cstdlib> #include <cmath> #include <cctype> #define Max(a,b) ((a)>(b)?(a):(b)) #define ls rt<<1 #define rs rt<<1|1 using namespace std; const int N = 200010; int xx[N];//x[i]用来储存海报左右边，排列后节点的位置为i; int post[N * 50];//post[i]用来储存边i应该排列在第几个位置上面 struct Poster { int l, r; }P[N];//P[i]来储存第i张海报的左右边 struct node { int l, r; bool cover; int Mid() { return (l + r) >> 1; } }a[N<<2]; int cmp(const void *a, const void *b) { int *c, *d; c = (int *)a; d = (int *)b; return *c - *d; } void Build(int rt, int l, int r) { a[rt].l = l, a[rt].r = r, a[rt].cover = false; if(l == r) return ; Build(ls, l, a[rt].Mid()); Build(rs, a[rt].Mid()+1, r); } void UP(int rt) { //如果ls和rs都被覆盖， 则此区域被覆盖 if(a[rt].l == a[rt].r) return; if(a[ls].cover && a[rs].cover) a[rt].cover = true; } bool Search(int rt, int l, int r)//如果找到此区间未被覆盖，覆盖并返回true, 否则返回false; { bool ans; if(a[rt].cover) return false; if(a[rt].l == l && a[rt].r == r) { if(a[rt].cover) return false; else { a[rt].cover = true; return true; } } if(l > a[rt].Mid()) { ans = Search(rs, l, r); } else if(r <= a[rt].Mid()) { ans = Search(ls, l, r); } else { bool ans1 = Search(ls, l, a[rt].Mid()); bool ans2 = Search(rs, a[rt].Mid() + 1, r); ans = ans1 || ans2; } UP(rt);//每次放置海报之后就要向上更新 return ans; } int main() { int T, n; scanf("%d", &T); while (T--) { scanf("%d", &n); int index = 0; for(int i = 0; i < n; i++) { scanf("%d %d", &P[i].l, &P[i].r); xx[index++] = P[i].l; xx[index++] = P[i].r; } qsort(xx, index, sizeof(xx[0]), cmp);//对每个左右边的大小排序，确定其属于的节点位置 int index2 = 1; for(int i = 0; i < index; i++) {//每个边的位置确定一下，相同值为一个节点，相邻节点也相邻，不相邻节点之间空一个节点 post[xx[i]] = index2; if(i == index - 1) break;//判断到最后一个，结束循环，因为每次循环保存的都是下个节点的状态，及时结束防止数组越界 if(xx[i + 1] - xx[i] == 1) index2++; if(xx[i + 1] - xx[i] > 1) index2 += 2; } Build(1, 1, index2);//建立线段树 int ans = 0;//保存答案 for(int i = n - 1; i >= 0; i--) { if(Search(1, post[P[i].l], post[P[i].r])) ans++;//每次贴一张海报，判断会不会被覆盖，不会则答案+1 } printf("%d\n", ans); } } /* 1 5 1 4 2 6 8 10 3 4 7 10 */
