LC 面试题 08.13. Pile Box LCCI

link

 题解：

按高度升序排序，若高度相同，按宽度降序排序。到i后，用线段树找w,d 在[0,0]-[wi-1,di-1]的最大值，更新此节点。

class Solution {
public:
    unordered_map<int,int> seg[3010<<2];
    int qx1,qy1,qx2,qy2;
    int x,y,val;
    int n;
    int cur;
    int queryx(int idx, int xleft, int xright){
        if(xleft>xright || qx1>xright || qx2<xleft) return 0;
        if(qx1<=xleft && qx2>=xright){
            int res=queryy(idx,0,0,n);
            return res;
        }

        int mid=(xleft+xright)/2;
        int l=queryx(idx*2+1,xleft,mid);
        int r=queryx(idx*2+2,mid+1,xright);
        return max(l,r);
    }
    int queryy(int idx, int idy, int yleft, int yright){
        if(qy1>yright || qy2<yleft) return 0;
        if(qy1<=yleft && qy2>=yright) return seg[idx][idy];

        int mid=(yleft+yright)/2;
        int l=queryy(idx,idy*2+1,yleft,mid);
        int r=queryy(idx,idy*2+2,mid+1,yright);
        return max(l,r);
    }

    void updatex(int idx, int xleft, int xright){
        if(xleft>xright || x<xleft || x>xright) return;
        if(xleft==xright){
            updatey(idx,xleft,xright,0,0,n);
            return;
        }
        int mid=(xleft+xright)/2;
        updatex(idx*2+1,xleft,mid);
        updatex(idx*2+2,mid+1,xright);
        updatey(idx,xleft,xright,0,0,n);
    }

    void updatey(int idx, int xleft, int xright, int idy, int yleft, int yright){
        if(yleft>yright || y<yleft || y>yright) return;
        if(yleft==yright){
            if(xleft==xright) {
                seg[idx][idy]=val;
            }
            else seg[idx][idy]=max(seg[idx*2+1][idy],seg[idx*2+2][idy]);
            return;
        }
        int mid=(yleft+yright)/2;
        updatey(idx,xleft,xright,idy*2+1,yleft,mid);
        updatey(idx,xleft,xright,idy*2+2,mid+1,yright);
        seg[idx][idy]=max(seg[idx][idy*2+1],seg[idx][idy*2+2]);
    }

    int pileBox(vector<vector<int>>& box) {
        unordered_map<int,int> wid;
        unordered_map<int,int> did;
        sort(box.begin(),box.end(),[](vector<int>& v1, vector<int>& v2){
            return v1[0]<v2[0];
        });
        n=box.size();
        for(int i=0;i<n;i++){
            if(i==0) wid[box[i][0]]=1;
            else {
                if(box[i][0]>box[i-1][0]) wid[box[i][0]]=i+1;
            }
        }
        sort(box.begin(),box.end(),[](vector<int>& v1, vector<int>& v2){
            return v1[1]<v2[1];
        });
        for(int i=0;i<n;i++){
            if(i==0) did[box[i][1]]=1;
            else {
                if(box[i][1]>box[i-1][1]) did[box[i][1]]=i+1;
            }
        }
        sort(box.begin(),box.end(),[](vector<int>& v1, vector<int>& v2){
            return v1[2]==v2[2]?v1[0]>v2[0]: v1[2]<v2[2];
        });
      
        int res=0;
        vector<int> dp(n,0);
        for(int i=0;i<n;i++){
            qx1=0;
            qy1=0;
            qx2=wid[box[i][0]]-1;
            qy2=did[box[i][1]]-1;
            int q=queryx(0,0,n);
            dp[i]=q+box[i][2];
            res=max(res,dp[i]);
            x=wid[box[i][0]];
            y=did[box[i][1]];
            val=dp[i];
            cur=i;
            updatex(0,0,n);
        }
        return res;
    }

   
};