hdoj 6047( 2017 Multi-University Training Contest

题目链接：Maximum Sequence

题目大意：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}

题目思路：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],所以我们可以有两种方法维护最大值：线段树或者优先队列，解法见代码

线段树：

#include <map>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1e9+7;int n;int ma[2000009];int b[250009];void build(int k, int l, int r){    if (l == r)    {        if (l > n) return;        scanf("%d", &ma[k]); ma[k] -= l;        return;    }    int mid = (l + r) >> 1;    build(k << 1, l, mid);    build(k << 1 | 1, mid + 1, r);    ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}int query(int k, int l, int r, int ll, int rr){    if (l >= ll&&r <= rr) return ma[k];    int ma = -1, mid = (l + r) >> 1;    if (mid >= ll) ma = max(ma, query(k << 1, l, mid, ll, rr));    if (rr > mid) ma = max(ma, query(k << 1 | 1, mid + 1, r, ll, rr));    return ma;}void update(int k, int l, int r, int p, int val){    if (l == r) { ma[k] = val; return; }    int mid = (l + r) >> 1;    if (p <= mid) update(k << 1, l, mid, p, val);    else update(k << 1 | 1, mid + 1, r, p, val);    ma[k] = max(ma[k << 1], ma[k << 1 | 1]);}int main(){    while(~scanf("%d",&n)){        build(1,1,2*n);        for(int i = 1;i <= n;i++)            scanf("%d",&b[i]);        sort(b+1,b+1+n);        int sum = 0;        for(int i = n+1;i <= n*2;i++){            int k = query(1,1,2*n,b[i-n],i-1);            //cout<<"k = "<<k<<endl;            update(1,1,2*n,i,k-i);            sum += k;            sum %= mod;        }        printf("%d\n",sum);    }    return 0;}

优先队列：

#include <map>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1e9+7;const int maxn = 3e5+10;int a[maxn],b[maxn],n;struct node{    int val,id;};struct cmp{    bool operator()(const node &a,const node &b){        return a.val < b.val;    }};priority_queue<node,vector<node>,cmp> Q;int main(){    while(~scanf("%d",&n)){        while(!Q.empty()) Q.pop();        for(int i = 1;i <= n;i++){            scanf("%d",&a[i]);            a[i] -= i;            Q.push((node){a[i],i});        }        for(int i = 1;i <= n;i++)            scanf("%d",&b[i]);        sort(b+1,b+1+n);        int sum = 0;        for(int i = n+1;i <= n*2;i++){            while(Q.top().id < b[i-n]) Q.pop();            sum += Q.top().val;            sum %= mod;            Q.push((node){Q.top().val-i,i});        }        printf("%d\n",sum);    }    return 0;}