Time Limit: 3 Seconds Memory Limit: 32768 KB
Now the God is very angry, so he wants to punish the lazy, greedy humans. He chooses to throw some lines of bricks (just down from the very high Heaven). These days the God lives in a 2D world, so he just throw the bricks in a vertical plane. Each time, the
God throws a line of bricks. The width of each brick is 1, and the length will be given.
t__nt is a hero in the world and he is trying his best to save the world. Now he has made m horizontal boards in the air with his magic power to stop the bricks. If one brick falls onto a board, it can not fall down any more. Notice that, for a line of bricks,
consecutive bricks are not connected. So when some bricks touch a board, the others will continues to fall down. Now, t__nt wants to know how many bricks each board holds after the God‘s crazy action. He asks you, an ACMer, to help him.
Input
There are no more then 10 cases. There is a blank line between consecutive cases. The first line of each case contains two integers n, m (0 < n, m <= 100000), indicating the number of lines of bricks and number of horizontal boards made by t__nt. n lines follow,
each contains two integers li, ri (0 <= li < ri <= 30000000). li and ri is the x-coordinates for the left side and the right side of the line of bricks. m lines follow, each contains three integers ai, bi, and hi (0 <= ai < bi <= 30000000; 0 < hi < 1000000000),
which means that board i is at height hi and the extreme points are ai and bi. You may assume no two boards with same height will overlap with each other.
Output
For each case, print m lines. The ith line means the number of bricks on board i at last. Print a blank line after each case.
Sample Input
1 2
1 8
1 5 8
3 7 6
Sample Output
4
2
这一题是the simple problem of integers的升级版，错了很多次。一开始我的思路是先放拦截的木板（木板先从低到高排序，因为高的会覆盖低的），然后依次读入木块，碰到混合的木块就继续向子树搜索，碰到同种颜色的木块就把加到这种颜色上，然后返回。但是这样如果颜色分的很杂的话，就会超时。所以我换了一种思路，先更新下落的木块，维护线段树的cnt和sum，cnt是延迟标志，表示这整段都要增加的长度，sum表示这段总的木块数（注意：这里的sum是实际的总覆盖数，也是加上cnt的覆盖数，即每次更新cnt的时候，sum也要更新，这里和the
simple problem of integers，因为那题cnt更新的时候可以不用更新sum,只有pushdown的时候才要把cnt变为0，sum+=cnt*(r-l+1).但是这一题多了一个清0操作，即要使这一段的数都变为0，如果不记录实际长度，就会出错，这里sum[i]=sum[i*2]+sum[i*2+1]是重点.）然后这题很坑的地方就是内存不足，我MLE了10多次，最后把long long开成int，然后把线段树的左右标记去掉，写进函数，终于A了。= =
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define maxn 100010
int pos[4*maxn];
int a[maxn],d[maxn];
struct edge{
int l,r,h,id;
ll num1;
}c[maxn];
ll sum[16*maxn],cnt[16*maxn];
bool flag[16*maxn];
bool cmp1(edge a,edge b){
return a.h>b.h;
}
bool cmp2(edge a,edge b){
return a.id<b.id;
}
void build(int l,int r,int i)
{
int mid;
cnt[i]=sum[i]=0;flag[i]=false;
if(l==r)return;
mid=(l+r)/2;
build(l,mid,i*2);
build(mid+1,r,i*2+1);
}
void update1(int l,int r,int L,int R,int i)
{
int mid;
if(L==l && R==r){
cnt[i]+=1;
sum[i]+=(ll)(pos[r+1]-pos[l]);
return;
}
mid=(L+R)/2;
if(cnt[i]){
cnt[i*2]+=cnt[i];
cnt[i*2+1]+=cnt[i];
sum[i*2]+=cnt[i]*(ll)(pos[mid+1]-pos[L]);
sum[i*2+1]+=cnt[i]*(ll)(pos[R+1]-pos[mid+1]);
cnt[i]=0;
}
if(r<=mid)update1(l,r,L,mid,i*2);
else if(l>mid)update1(l,r,mid+1,R,i*2+1);
else{
update1(l,mid,L,mid,i*2);
update1(mid+1,r,mid+1,R,i*2+1);
}
sum[i]=sum[i*2]+sum[i*2+1];
}
void update2(int l,int r,int color,int L,int R,int i)
{
int mid;
if(flag[i])return;
if(l==L && R==r){
flag[i]=true;
c[color].num1+=sum[i];
sum[i]=0;
return;
}
mid=(L+R)/2;
if(cnt[i]){
cnt[i*2]+=cnt[i];
cnt[i*2+1]+=cnt[i];
sum[i*2]+=cnt[i]*(ll)(pos[mid+1]-pos[L]);
sum[i*2+1]+=cnt[i]*(ll)(pos[R+1]-pos[mid+1]);
cnt[i]=0;
}
if(r<=mid)update2(l,r,color,L,mid,i*2);
else if(l>mid)update2(l,r,color,mid+1,R,i*2+1);
else{
update2(l,mid,color,L,mid,i*2);
update2(mid+1,r,color,mid+1,R,i*2+1);
}
sum[i]=sum[i*2]+sum[i*2+1];
}
int main()
{
int n,m,i,j,t,tot,t1,t2;
while(scanf("%d%d",&n,&m)!=EOF)
{
t=0;
for(i=1;i<=n;i++){
scanf("%d%d",&a[i],&d[i]);
t++;pos[t]=a[i];
t++;pos[t]=d[i];
}
for(i=1;i<=m;i++){
scanf("%d%d%d",&c[i].l,&c[i].r,&c[i].h);
c[i].id=i;c[i].num1=0;
t++;pos[t]=c[i].l;
t++;pos[t]=c[i].r;
}
sort(pos+1,pos+1+t);
tot=1;
for(i=2;i<=t;i++){
if(pos[i]!=pos[tot]){
tot++;pos[tot]=pos[i];
}
}
build(1,tot-1,1);
for(i=1;i<=n;i++){
t1=lower_bound(pos+1,pos+1+tot,a[i])-pos;
t2=lower_bound(pos+1,pos+1+tot,d[i])-pos;
update1(t1,t2-1,1,tot-1,1);
}
sort(c+1,c+1+m,cmp1);
for(i=1;i<=m;i++){
t1=lower_bound(pos+1,pos+1+tot,c[i].l)-pos;
t2=lower_bound(pos+1,pos+1+tot,c[i].r)-pos;
update2(t1,t2-1,c[i].id,1,tot-1,1);
}
for(i=1;i<=m;i++){
printf("%lld\n",c[i].num1);
}
printf("\n");
}
return 0;
}
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