ZOJ 3299 线段树

//题意：有n排板砖，m个木板，边界的l和r的n列板砖从天上掉下来，然后有m个边界的a,b的高度为h的木板去接那些板砖，一排板砖中的部分板砖如果掉到木板上就停止下落，//剩下的继续下落，问最后每块木板上有多少块板砖。////开结构体去存储线段树中节点的信息MLE了。。////按木板的高度，从低到高，去更新线段树节点对应的木板的编号。////板砖落下时，将对应区间的覆盖次数加一。////最后查询的时候，如果已经是叶子结点，或者这个区间的对应的木板的编号是确定的，那么就要在对应的木板编号上加上，覆盖次数乘以区间长度的值。//////all is for 普吉岛*******************************************************************************************////就是我要好好的敲一下离散化////先是更新一遍,高的模板覆盖低的模板，这样的话就变成一条木板了；//然后再用木块覆盖上去，更新每一段的覆盖次数；//最后查询每一块的覆盖次数*区间长度。。//首先离散化。。////想到最好要养成代码写完就能一遍过的情况，不要每次都在那里等调试。//还有一个很重要的问题，做题的速度问题！！！//一开始为什么会mle?过了//传参？map?vector?re?#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<algorithm>#include<string.h>#define maxn 100010#include<vector>#include<map>#define ll long longusing namespace std;#define maxm 1600100#define rep(i,n) for(int i=0;i<n;++i)map<int,int>H;vector<int> Y;//where is the problemstruct node{int sum,l,r,col;}tree[maxm];long long ans[maxn];int LL[401000];void build(int root ,int left,int right){//    cout<<left<<" "<<right<<" "<<root<<endl;    tree[root].col=-1;    tree[root].sum=0;    tree[root].l=left;    tree[root].r=right;    if(left==right-1) return ;    else{    int mid=(left+right)>>1;    build(root<<1,left,mid);    build(root<<1|1,mid,right);}}void updatec(int root,int s,int t,int cl){//    cout<<root<<" "<<s<<" "<<t<<endl;    int left=tree[root].l;    int right=tree[root].r;//    cout<<LL[left]<<" "<<LL[right]<<" "<<s<<" "<<t<<endl;    if(LL[left]>=s&& LL[right]<=t) {tree[root].col=cl;return;}//    cout<<LL[left]<<" "<<LL[right]<<" "<<cl<<endl;return;}    if(left+1==right) return;    if(tree[root].col != -1)    {    tree[root<<1|1].col=tree[root].col;    tree[root<<1].col=tree[root].col;    tree[root].col=-1;    }    int mid=(left+right)>>1;    if(s<=LL[mid]) updatec(root<<1,s,t,cl);    if(t>=LL[mid]) updatec(root<<1|1,s,t,cl);}void updateS(int root,int s,int t){//    cout<<root<<" "<<s<<" "<<t<<endl;    int left=tree[root].l;    int right=tree[root].r;    if(LL[left]>=s &&LL[right]<=t)    {        tree[root].sum++;        return;    }    if(left+1==right) return;    if(tree[root].sum>0)    {       tree[root<<1].sum+=tree[root].sum;        tree[root<<1|1].sum+=tree[root].sum;        tree[root].sum=0;    }    int mid=(left+right)>>1;    if(s<=LL[mid]) updateS(root<<1,s,t);    if(t>=LL[mid]) updateS(root<<1|1,s,t);}void query(int root){//    cout<<root<<endl;    int left=tree[root].l;    int right=tree[root].r;    if(left+1 == right){if(tree[root].col != -1) ans[tree[root].col] += (long long)(LL[tree[root].r] - LL[tree[root].l]) * (long long )tree[root].sum;return ;}    if(tree[root].sum > 0){tree[root<<1].sum += tree[root].sum;tree[root<<1|1].sum += tree[root].sum;tree[root].sum = 0;}if(tree[root].col != -1){tree[root<<1].col= tree[root<<1|1].col = tree[root].col;}int mid = (left + right) >> 1;query(root<<1);query(root<<1|1);}void print(int root){//    cout<<tree[root].l<<" "<<tree[root].r<<" "<<tree[root].sum<<" "<<tree[root].col<<endl;    if(tree[root].l==tree[root].r-1) return;    print(root<<1);    print(root<<1|1);}struct Node{    int l,r,h,index;}boa[maxn];int l[maxn];int r[maxn];bool cmp(Node a,Node b){    return a.h<b.h;}int pos;struct Plane{    int x,y;}plane[maxn];struct Block{    int x,y,z;int idx;}block[maxn];bool cmp1(Block a,Block b){    return a.z < b.z;}int main() {//freopen("in.txt","r",stdin);int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        pos = 0;        rep(i,n){            scanf("%lld%lld",&plane[i].x,&plane[i].y);            LL[pos++] = plane[i].x;            LL[pos++] = plane[i].y;        }        rep(i,m){            scanf("%d%d%d",&block[i].x,&block[i].y,&block[i].z);            block[i].idx = i+1;           LL[pos++] = block[i].x;            LL[pos++] = block[i].y;        }        sort(block,block+m,cmp1);        sort(LL,LL+pos);        int len = unique(LL,LL+pos)-LL-1;        build(1,0,len);        rep(i,m){            updatec(1,block[i].x,block[i].y,block[i].idx);        }        memset(ans,0,sizeof(ans));        rep(i,n){            updateS(1,plane[i].x,plane[i].y);        }        query(1);//        print(1);        for(int i=1;i<=m;++i){         printf("%lld\n",ans[i]);        }        printf("\n");    }return 0;}