[Codeforces 489E] Nastya and King-Shamans

[题目链接]

        http://codeforces.com/contest/992/problem/E

[算法]

        线段树 + 二分

         时间复杂度 : O(NlogN^2)

[代码]

        

#include
using namespace std;
const int MAXN = 2e5 + 10;
typedef long long ll;

struct Node
{
    int l,r;
    ll mx,sum;
} Tree[MAXN << 2];

int i,n,q,x,y,cur,tmp,ans;
ll value[MAXN];
ll pre;

template  inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
    {
        if (c == '-') f = -f;
    }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void update(int index)
{
    Tree[index].mx = max(Tree[index << 1].mx,Tree[index << 1 | 1].mx);
    Tree[index].sum = Tree[index << 1].sum + Tree[index << 1 | 1].sum;
}
inline void build(int index,int l,int r)
{
    int mid;
    Tree[index].l = l;
    Tree[index].r = r;
    if (l == r)
    {
        Tree[index].mx = value[l];
        Tree[index].sum = value[l];
        return;
    }
    mid = (l + r) >> 1;
    build(index << 1,l,mid);
    build(index << 1 | 1,mid + 1,r);
    update(index);
}
inline void modify(int index,int pos,int val)
{
    int mid;
    if (Tree[index].l == Tree[index].r)
    {
        Tree[index].mx = Tree[index].sum = val;
        return;
    }
    mid = (Tree[index].l + Tree[index].r) >> 1;
    if (mid >= pos) modify(index << 1,pos,val);
    else modify(index << 1 | 1,pos,val);
    update(index);
}
inline int query(int index,int l,int r,ll val)
{
    int mid,tmp;
    if (Tree[index].l == l && Tree[index].r == r)
    {
        if (Tree[index].mx < val) return -1;
        if (l == r) return l;
        mid = (Tree[index].l + Tree[index].r) >> 1;
        if (Tree[index << 1].mx >= val) return query(index << 1,l,mid,val);
        else return query(index << 1 | 1,mid + 1,r,val);    
    }    
    mid = (Tree[index].l + Tree[index].r) >> 1;
    if (mid >= r) tmp = query(index << 1,l,r,val);
    else if (mid + 1 <= l) tmp = query(index << 1 | 1,l,r,val);
    else 
    {
        tmp = query(index << 1,l,mid,val);
        if (tmp != -1) return tmp;
        return query(index << 1 | 1,mid + 1,r,val);
    }
    return tmp;
}
inline ll query_sum(int index,int l,int r)
{
    int mid;
    if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
    mid = (Tree[index].l + Tree[index].r) >> 1;
    if (mid >= r) return query_sum(index << 1,l,r);
    else if (mid + 1 <= l) return query_sum(index << 1 | 1,l,r);
    else return query_sum(index << 1,l,mid) + query_sum(index << 1 | 1,mid + 1,r);
}
int main()
{
    
    read(n); read(q);
    for (i = 1; i <= n; i++) read(value[i]);
    build(1,1,n);
    while (q--)
    {
        read(x); read(y);
        value[x] = y;
        modify(1,x,y);
        if (value[1] == 0)
        {
            printf("1\n");
            continue;    
        }    
        cur = pre = tmp = 0; ans = -1;
        while (cur < n)
        {
            tmp = query(1,cur + 1,n,pre);
            if (tmp == -1) break;
            cur = tmp;
            pre = query_sum(1,1,tmp);
            if (pre - value[cur] == value[cur]) 
            {
                ans = tmp;
                break;
            }
        }
        printf("%d\n",ans);
    }
    
    return 0;
}

 

转载于:https://www.cnblogs.com/evenbao/p/9487674.html