HDU 6521 Party（线段树）

　　　题目意思：有n个人，一开始相互不认识。他们要去参加party，每次参加的人是编号在区间[l,r]内的人。参加完一次party之后，这区间内的人就会相互认识。每次会有有多少对人会新认识。那么用f[i] = j 表示 j - i的人都互相认识了，初始化f[i] = i,那么每次更新l,r话只需要找到是否存在f[l - r] > l并进行更新。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e5 + 10;

int n, m;
int val[N << 2], num[N  << 2];

int read()
{
    int res = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9')
    {
        if(c == -1)
        {
            f = -1;
        }
        c = getchar(); 
    }
    while(c >= '0' && c <= '9')
    {
        res = res * 10 + c - '0';
        c = getchar();
    }
    return f * res;
}

void pushup(int rt)
{
    val[rt] = max(val[rt << 1], val[rt << 1 | 1]);
}

void build(int rt, int l, int r)
{
    if(l == r)
    {
        val[rt] = num[rt] = l;
        //puts("!");
        return ;
    }
    
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    pushup(rt);
} 

ll query(int L, int R, int l, int r, int rt, int x)
{
    if(x >= val[rt]) return 0;
    if(l == r)
    {
        ll re = num[rt] - x;
        val[rt] = num[rt] = x;
        return re;
    }
    if(L <= l && r <= R)
    {
        if(x >= val[rt]) return 0;
        else
        {
            ll re = 0;
            int mid = l + r >> 1;
            re += query(L, min(R, mid), l, mid, rt << 1, x);
            re += query(max(L, mid + 1), R, mid + 1, r, rt << 1 | 1, x);

            pushup(rt);
            return re;
        }
    }
    ll res = 0;
    int mid = l + r >> 1;
    if(L <= mid)
    res += query(L, min(R, mid), l, mid, rt << 1, x);
    if(R > mid)
    res += query(max(L, mid + 1), R, mid + 1, r, rt << 1 | 1, x);
    pushup(rt);
    return res;
}


int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        build(1, 1, n);
        
    //    cout << val[1] <<" ! " << endl;
        int x, y;
        while(m --)
        {
        x = read(), y = read();
        ll ans = query(x, y, 1, n, 1, x);
        printf("%lld\n", ans);
        }
    }
}