标签：lse sub nes base ref ring ack mat into
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6638
InputThe first line of the input contains an integer T(1≤T≤100)T(1≤T≤100), denoting the number of test cases.
In each test case, there is one integer n(1≤n≤2000)n(1≤n≤2000) in the first line, denoting the number of pirate chests.
For the next nn lines, each line contains three integers xi,yi,wi(−109≤xi,yi,wi≤109)xi,yi,wi(−109≤xi,yi,wi≤109), denoting each pirate chest.
It is guaranteed that ∑n≤10000∑n≤10000.OutputFor each test case, print a single line containing an integer, denoting the maximum total value.Sample Input
2 4 1 1 50 2 1 50 1 2 50 2 2 -500 2 -1 1 5 -1 1 1
Sample Output
100 6
题目大意：二维坐标中有一些点，每个点有一个值，你有一个任意大小的矩形，问你最大能框出的值是多少
思路：点的个数在2000个以内，我们可以枚举矩形的左右边界，对于左右边界以内的点，用线段树维护区间最大值就行了
有个需要注意的地方就是既然要枚举边界，就要一次性枚举完，不然算出来的值肯定是不准确的
看代码：
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #include<map> #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #include<map> #include<queue> #include<cmath> using namespace std; typedef long long LL; typedef unsigned long long ull; #define sc1(a) scanf("%lld",&a) #define pf1(a) printf("%lld\n",a) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 const LL INF=1e18; const ull base=2333; const int maxn=2e3+50; const int maxm=1e3+50; const int maxv=1e6+5; const int mod=1e9+7; const int ba=3e5; struct Node { LL x,y,w; bool operator < (const Node &b) const { return b.x<x; } }a[maxn]; struct Tree { LL ll,rr,lm,sum;// }t[maxn<<2]; LL ly[maxn]; LL len; LL get_id(LL w) { return lower_bound(ly+1,ly+1+len,w)-(ly); } void Push_up(LL rt) { t[rt].ll=max(t[rt<<1].ll,t[rt<<1].sum+t[rt<<1|1].ll); t[rt].rr=max(t[rt<<1|1].rr,t[rt<<1|1].sum+t[rt<<1].rr); t[rt].lm=max(t[rt<<1].lm,max(t[rt<<1|1].lm,t[rt<<1].rr+t[rt<<1|1].ll)); t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum; } void Update(LL l,LL r,LL rt,LL L,LL v) { if(l==r&&l==L) { t[rt].lm+=v; t[rt].ll=t[rt].rr=t[rt].sum=t[rt].lm; return ; } LL mid=(l+r)>>1; if(L<=mid) Update(l,mid,rt<<1,L,v); else Update(mid+1,r,rt<<1|1,L,v); Push_up(rt); } int main() { LL T;sc1(T); while(T--) { LL ans=0; LL N;sc1(N); for(LL i=1;i<=N;i++) { sc1(a[i].x);sc1(a[i].y);sc1(a[i].w); ly[i]=a[i].y;//离散化y } sort(a+1,a+1+N); sort(ly+1,ly+1+N); len=unique(ly+1,ly+1+N)-(ly+1); for(LL i=1;i<=N;i++) //枚举矩形左边界 { if(i!=1&&a[i].x==a[i-1].x) continue; for(LL j=1;j<=(len<<2);j++) t[j].ll=t[j].rr=t[j].sum=t[j].lm=0; for(LL j=i;j<=N;j++) //枚举矩形右边界 { LL v=get_id(a[j].y); Update(1,len,1,v,a[j].w); if(j==N||a[j].x!=a[j+1].x)//这里是一个需要注意的点 我们必须对于每一条边界 一定要枚举完 ans=max(ans,t[1].lm); } } pf1(ans); } return 0; } /** */
标签：lse sub nes base ref ring ack mat into
原文地址：https://www.cnblogs.com/caijiaming/p/12467862.html
