题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6047
题目：
Maximum Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 90 Accepted Submission(s): 44
Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .
Now Steph finds it too hard to solve the problem, please help him.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.
多校联赛第二场~
题意：
给定一个长度为n的a数组和b数组，要求a[n+1]…a[2*n]的最大总和。 限制条件为ai≤max{aj-j│bk≤j<i}。
思路：
a[j](j>n)是从当前选择的a数组的b[k]个数开始，到最后一个数中选。由于每个b[k]都只能使用一次，我们要可能地把b[k]较大的数留在后面用，因为刚开始a数组只有n个，只有随着每次操作a数组才会增加一个数。
顺着这个思路，我们很自然地先对b数组做一次升序排序，再以b[k]为左区间，a数组当前的个数为右区间，来找最大的a[j]; 因为数据量比较大，我们经常要获取某个区间a[j]的最大值，所以用线段树维护。
代码：
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int N=*+;
const int M= 1e9+;
typedef long long ll;
struct node{
int l,r;
int Max;
ll sum;
}tree[*N];
int n;
int a[N],b[N];
void pushup(int i){
tree[i].sum=(tree[*i].sum+tree[*i+].sum)%M;//别忘了mod运算
tree[i].Max=max(tree[*i].Max,tree[*i+].Max);
}
void build(int l,int r,int i){
if(i>*n) return ;
tree[i].l=l;
tree[i].r=r;
if(tree[i].l==tree[i].r) {
if(l>n){
tree[i].sum=;
tree[i].Max=;
}else{
tree[i].sum=a[l];
tree[i].Max=a[l]-l;//MAX存的是a[j]-j;
}
return ;
}
int mid=(l+r)/;
build(l,mid,*i);
build(mid+,r,*i+);
pushup(i);//回溯更新父节点
}
void update(ll v,int x,int i){
if(tree[i].l==tree[i].r){
tree[i].sum=v;
tree[i].Max=v-x;
return ;
}
int mid=(tree[i].l+tree[i].r)/;
if(x<=mid) update(v,x,*i);
else update(v,x,*i+);
pushup(i);
}
int query(int l,int r,int i){
if(tree[i].l==l && tree[i].r==r) return tree[i].Max;
int mid=(tree[i].l+tree[i].r)/;
if(r<=mid) return query(l,r,*i);
else if(l>mid) return query(l,r,*i+);
else if(l<=mid && r>mid) return max(query(l,mid,*i),query(mid+,r,*i+));
return -;
}
int main(){
while(scanf("%d",&n)!=EOF){
int pre=;
for(int i=;i<=n;i++) scanf("%d",&a[i]);
for(int i=;i<n;i++) scanf("%d",&b[i]);
build(,*n,);
sort(b,b+n);
for(int i=n+;i<=*n;i++){
int x,y;
int bg,ed=i-;
x=bg=b[pre++];//排序完直接按顺序取b数组，保证了不会重复使用
y=query(bg,ed,);
a[i]=max(x,y);
update(a[i],i,);
}
printf("%lld\n",tree[].sum);//tree[3]保存的是n+1…2*n的节点信息
}
return ;
}