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最开始觉得和zoj1610 count colors应该是一个模板
但是交了以后发现MLE
数据范围太大 没办法直接用数组 所以要离散化
第一次用到离散化 学了个新内容 代码是借鉴了题解
eg 范围[1,6] [1.7] [2,10] [8 18] 将各点排序
1 1 2 6 7 8 10 18 离散后对应的坐标为
1 2 3 4 5 6 7 再根据原来的点把它们对应起来，则离散后坐标为
[1,3] [1,4] [2,6] [5,7]
离散化代码
for(int i = 0; i < n * 2; i += 2){
//int a, b;
scanf("%d%d",&post[i].x,&post[i + 1].x);
post[i].id = post[i + 1].id = i;
}
sort(post, post + 2 * n, cmp1);
int tot = 0, pre = 0;
for(int i = 0; i < 2 * n; i++){
if(post[i].x == pre)
post[i].x = tot;
else{
pre = post[i].x;
post[i].x = ++tot;
}
}从后往前贴 能保证已经能看见的poster里总能有最后的一部分不会被遮住
如果发现这段区域已经被完全覆盖了就返回
完整代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<limits>
#include<stack>
#include<queue>
#include<cmath>
#define inf 1000005
//Ï߶ÎÊ÷DÌâ
//http://blog.csdn.net/dt2131/article/details/52919953
//http://www.cnblogs.com/jackge/archive/2013/04/25/3041637.html
using namespace std;
const int maxn = 100010;
int c, n,flag;
//int tree[maxn << 2], dis[maxn];
bool vis[maxn];
struct node{
int id, x;
}post[maxn << 2];
struct Tree{
int l, r;
bool vis;
}tree[maxn << 2];
void build(int L, int R, int rt)
{
tree[rt].l = L;
tree[rt].r = R;
tree[rt].vis = 0;
if(tree[rt].l == tree[rt].r)
return;
int mid = (L + R) / 2;
build(L, mid, rt * 2);
build(mid + 1, R, rt * 2 + 1);
}
void pushup(int rt)
{
tree[rt].vis = tree[2 * rt].vis && tree[2 * rt + 1].vis;
}
/*void pushdown(int rt)
{
if(tree[rt] != -1){
tree[2 * rt] = tree[2 * rt + 1] = tree[rt];
tree[rt] = -1;
}
}
void update(int a, int b, int data, int l, int r, int rt)
{
if(a <= l && b >= r){
tree[rt] = data;
return;
}
if(tree[rt] == data)
return;
pushdown(rt);
int mid = (l + r) / 2;
if(a <= mid)
update(a, b, data, l, mid, 2 * rt);
if(b > mid)
update(a, b, data, mid + 1, r, 2 * rt + 1);
}*/
/*void query(int l, int r, int rt)
{
if(tree[rt] != -1){
for(int i = l; i <= r; i++){
dis[i] = tree[rt];
}
return;
}
if(l != r && tree[rt] == -1){
int mid = (l + r) / 2;
query(l, mid, 2 * rt);
query(mid + 1, r, 2 * rt + 1);
}
}
*/
void query(int L, int R, int rt)
{
if(tree[rt].vis)
return;
if(tree[rt].l == L && tree[rt].r == R)
{
tree[rt].vis = 1;
flag = 1;
return;
}
int mid = (tree[rt].l + tree[rt].r) / 2;
if(R <= mid)
query(L, R, 2 * rt);
else if(L >= mid + 1)
query(L, R, 2 * rt + 1);
else{
query(L, mid, 2 * rt);
query(mid + 1, R, 2 * rt + 1);
}
pushup(rt);
}
bool cmp1(node a, node b)
{
return a.x < b.x;
}
bool cmp2(node a, node b)
{
if(a.id == b.id)
return a.x < b.x;
return a.id > b.id;
}
int main()
{
scanf("%d",&c);
while(c--){
scanf("%d",&n);
//memset(tree, -1, sizeof(tree));
//memset(dis, -1, sizeof(dis));
//int k = 1;
for(int i = 0; i < n * 2; i += 2){
//int a, b;
scanf("%d%d",&post[i].x,&post[i + 1].x);
post[i].id = post[i + 1].id = i;
}
sort(post, post + 2 * n, cmp1);
int tot = 0, pre = 0;
for(int i = 0; i < 2 * n; i++){
if(post[i].x == pre)
post[i].x = tot;
else{
pre = post[i].x;
post[i].x = ++tot;
}
}
build(1, 2 * n, 1);
sort(post, post + 2 * n, cmp2);
int ans = 0;
for(int i = 0; i < 2 * n; i += 2){
int l = post[i].x;
int r = post[i + 1].x;
flag = 0;
query(l, r, 1);
if(flag)
ans++;
}
/*memset(vis, 0, sizeof(vis));
int res = 0;
for(int i = 0; i < maxn;){
while(i < maxn && dis[i] == -1)
i++;
if(i >= maxn)
break;
int temp = dis[i];
if(!vis[temp]){
vis[temp] = true;
res++;
}
//res[temp]++;
while(i < maxn && dis[i] == temp)
i++;
}
/*int ans = 0;
for(int i = 0; i < maxn; i++){
ans += res[i];
}*/
printf("%d
",ans);
}
return 0;
}