This problem is more than Hangzhou electric school sixth title of 2019
Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=6638
Meaning of the questions: give you n points in the plane, each point has the right value (negative rights), so you might calculate a matrix of maximum coverage and weights;
Ideas: The maximum continuous sub-segment - segment tree bound enumerate, line by line by line updating of the segment tree point interpolation to one row is updated every answer (similar enumeration bounds), the time complexity: O (n ^ 2 * log (n));
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream> 6 #include <bitset> 7 #include <map> 8 //#include<unordered_map> http://acm.hdu.edu.cn/showproblem.php?pid=6638 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 //#include <windows.h>//reverse(arr,arr+len);// ~ ! ~ ! floor 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 22 using namespace std;//next_permutation(arr+1,arr+1+n);//prev_permutation 23 #define fo(arr,b,c) for(register int arr=b;arr<=c;++arr) 24 #define fr(arr,b,c) for(register int arr=b;arr>=c;--arr) 25 #define mem(arr,b) memset(arr,b,sizeof(arr)) 26 #define pr printf 27 #define sc scanf 28 #define ls rt<<1 29 #define rs rt<<1|1 30 void swapp(int &arr,int &b); 31 double fabss(double arr); 32 int maxx(int arr,int b); 33 int minn(int arr,int b); 34 int Del_bit_1(int n); 35 int lowbit(int n); 36 int abss(int arr); 37 //const long long INF=(1LL<<60); 38 const double E=2.718281828; 39 const double PI=acos(-1.0); 40 const int inf=(1<<29); 41 const double ESP=1e-9; 42 const int mod=(int)1e9+7; 43 const int N=(int)2003; 44 45 int b[N]; 46 struct node_ 47 { 48 int x,y; 49 long long v; 50 friend bool operator<(node_ a,node_ b) 51 { 52 if(a.y==b.y) 53 return a.x<b.x; 54 return a.y>b.y; 55 } 56 }arr[N]; 57 struct vnode 58 { 59 int x; 60 long long v; 61 }; 62 //=================================================离散化; 63 int LSx(int n) 64 { 65 int m=0; 66 for(int i=1;i<=n;++i) 67 b[++m]=arr[i].x; 68 sort(b+1,b+1+m); 69 m=unique(b+1,b+1+m)-b-1; 70 for(int i=1;i<=n;++i) 71 arr[i].x=lower_bound(b+1,b+1+m,arr[i].x)-b; 72 return m; 73 } 74 int LSy(int n) 75 { 76 int m=0; 77 for(int i=1;i<=n;++i) 78 b[++m]=arr[i].y; 79 sort(b+1,b+1+m); 80 m=unique(b+1,b+1+m)-b-1; 81 for(int i=1;i<=n;++i) 82 arr[i].y=lower_bound(b+1,b+1+m,arr[i].y)-b; 83 return m; 84 } 85 //===========================================连续子段线段树; 86 struct node 87 { 88 long long Sum,Lsum,Rsum,ans; 89 }tr[N<<2]; 90 91 void up(int rt) 92 { 93 tr[rt].Sum=tr[ls].Sum+tr[rs].Sum; 94 tr[rt].ans=max(max(tr[ls].ans,tr[rs].ans),tr[ls].Rsum+tr[rs].Lsum); 95 tr[rt].Lsum=max(tr[ls].Lsum,tr[ls].Sum+tr[rs].Lsum); 96 tr[rt].Rsum=max(tr[rs].Rsum,tr[rs].Sum+tr[ls].Rsum); 97 } 98 void Build(int l,int r,int rt) 99 { 100 if(l==r) 101 { 102 tr[rt].Sum=tr[rt].Lsum=tr[rt].Rsum=tr[rt].ans=0; 103 return; 104 } 105 int mid=(l+r)>>1; 106 Build(l,mid,ls); 107 Build(mid+1,r,rs); 108 up(rt); 109 } 110 void update_dot(int pos,long long v,int l,int r,int rt) 111 { 112 if(l==r) 113 { 114 tr[rt].Sum+=v; 115 tr[rt].Lsum=tr[rt].Rsum=tr[rt].ans=tr[rt].Sum; 116 return; 117 } 118 int mid=(l+r)>>1; 119 if(pos<=mid) 120 update_dot(pos,v,l,mid,ls); 121 else 122 update_dot(pos,v,mid+1,r,rs); 123 up(rt); 124 } 125 vector<vector<vnode> >v(N); 126 127 int main() 128 { 129 int T; 130 sc("%d",&T); 131 while(T--) 132 { 133 int n; 134 sc("%d",&n); 135 for(int i=1;i<=n;++i) 136 sc("%d%d%lld",&arr[i].x,&arr[i].y,&arr[i].v); 137 int L=LSx(n); 138 int H=LSy(n); 139 fo(i,1,H)v[i].clear(); 140 for(int i=1;i<=n;++i) 141 v[arr[i].y].push_back({arr[i].x,arr[i].v}); 142 long long ans=0; 143 for(int i=H;i>=1;--i) 144 { 145 Build(1,L,1); 146 for(int j=i;j>=1;--j) 147 { 148 int sz=v[j].size(); 149 for(int k=0;k<sz;++K) 150 { 151 the vnode T = V [J] [K]; 152 update_dot (TX, TV, . 1 , L, . 1 ); 153 } 154 ANS = max (ANS, TR [ . 1 ] .ans); // every an additional layer is updated answer; 155 } 156 } 157 PR ( " % LLD \ n- " , ANS); 158 } 159 return 0 ; 160. } 161 162 / **************************************************************************************/ 163 164 int maxx(int arr,int b) 165 { 166 return arr>b?arr:b; 167 } 168 169 void swapp(int &arr,int &b) 170 { 171 arr^=b^=arr^=b; 172 } 173 174 int lowbit(int n) 175 { 176 return n&(-n); 177 } 178 179 int Del_bit_1(int n) 180 { 181 return n&(n-1); 182 } 183 184 int abss(int arr) 185 { 186 return arr>0?arr:-arr; 187 } 188 189 double fabss(double arr) 190 { 191 return arr>0?arr:-arr; 192 } 193 194 int minn(int arr,int b) 195 { 196 return arr<b?arr:b; 197 }