http://acm.hdu.edu.cn/showproblem.php?pid=6638
题意：给你一些点的权值，让找一个矩形圈住一部分点，问圈住点的最大权值和
分析：由于是稀疏图，明显要先把x,y坐标离散化，暴力是n^3？(枚举边界n^2，求和是n)显然过不了，那可以枚举y的边界，然后对于x就是最大子段和的问题了，用线段树维护，n^2logn可以过。
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 4e3+5; 4 const int inf = 0x3f3f3f; 5 typedef pair<int,int> P; 6 typedef long long ll; 7 int cas,n; 8 struct point{ 9 int x,y,val; 10 bool operator<(const point &a) const{ 11 return y<a.y; 12 } 13 }a[maxn]; 14 vector<int> p[maxn]; 15 int bx[maxn],by[maxn]; 16 17 struct node{ 18 ll sum,lmax,rmax,lrmax; 19 }tree[maxn<<2]; 20 21 inline void pushup(int rt){ 22 tree[rt].sum = tree[rt<<1].sum+tree[rt<<1|1].sum; 23 tree[rt].lmax = max(tree[rt<<1].lmax,tree[rt<<1].sum+tree[rt<<1|1].lmax); 24 tree[rt].rmax = max(tree[rt<<1|1].rmax,tree[rt<<1].rmax+tree[rt<<1|1].sum); 25 tree[rt].lrmax = max(max(tree[rt<<1].lrmax,tree[rt<<1|1].lrmax),tree[rt<<1].rmax+tree[rt<<1|1].lmax); 26 } 27 28 inline void update(int L,int l,int r,int rt,int c){ 29 if(l==r){ 30 tree[rt].sum += 1ll*c; 31 tree[rt].lrmax = tree[rt].lmax = tree[rt].rmax = tree[rt].sum; 32 return; 33 } 34 int mid = l+r>>1; 35 if(L<=mid) update(L,l,mid,rt<<1,c); 36 else update(L,mid+1,r,rt<<1|1,c); 37 pushup(rt); 38 } 39 40 int main(){ 41 cin>>cas; 42 while(cas--){ 43 cin>>n; 44 for(int i=1;i<=n;i++){ 45 cin>>a[i].x>>a[i].y>>a[i].val; 46 bx[i] = a[i].x,by[i] = a[i].y; 47 } 48 sort(bx+1,bx+1+n); 49 int xlen = unique(bx+1,bx+1+n)-bx-1; 50 sort(by+1,by+1+n); 51 int ylen = unique(by+1,by+1+n)-by-1; 52 for(int i=1;i<=n;i++){ 53 a[i].x = lower_bound(bx+1,bx+xlen+1,a[i].x)-bx; 54 a[i].y = lower_bound(by+1,by+ylen+1,a[i].y)-by; 55 } 56 sort(a+1,a+1+n); 57 ll ans = 0; 58 for(int i=1;i<=ylen;i++){ 59 memset(tree,0,(xlen*4+5)*sizeof(node)); 60 int pos = 1; 61 while(a[pos].y<i&&pos<=n) pos++; 62 for(int j=i;j<=ylen;j++){ 63 while(a[pos].y<=j&&pos<=n){ 64 update(a[pos].x,1,xlen,1,a[pos].val); 65 pos++; 66 } 67 ans = max(ans,tree[1].lrmax); 68 } 69 70 } 71 cout<<ans<<endl; 72 } 73 return 0; 74 }