HDU 5741 Helter Skelter

离线处理+扫描线。题意很容易转化：若干个矩形形成并集，询问一些点是否在并集中？

官方题解不是这样做的....那种做法效率更高，暂时还不会。我这样是4500ms G++过的，C++TLE......

区间加上某值，询问单点值，可以用树状数组。用线段树可能常数较大导致TLE。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0), eps = 1e-;
void File()
{
    freopen("D:\\in.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
}
inline int read()
{
    char c = getchar();  while (!isdigit(c)) c = getchar();
    int x = ;
    while (isdigit(c)) { x = x *  + c - ''; c = getchar(); }
    return x;
}

int T, n, ps, pq, v[], c[ + ];
int A[ + ], sz;
struct Seg { int x, y1, y2, f; }s[ + ]; int ns;
struct Quary { int x, y, id; }q[ + ]; int nq;
int ans[ + ];

int lowbit(int x) { return x&(-x); }
void add(int p, int val) { while (p <= sz) c[p] = c[p] + val, p = p + lowbit(p); }
int sum(int p) { int r = ; while (p > ) r = r + c[p], p = p - lowbit(p); return r; }
void update(int L, int R, int val) { add(L, val); add(R + , -val); }

void AddSeg(int l, int r, int L, int R)
{
    s[ns].x = l, s[ns].y1 = L, s[ns].y2 = R, s[ns].f = , ns++;
    s[ns].x = r, s[ns].y1 = L, s[ns].y2 = R, s[ns].f = -, ns++;
    A[sz++] = L, A[sz++] = R;
}

bool cmp(Seg a, Seg b) { if (a.x == b.x) return a.f > b.f; return a.x < b.x; }
bool cmp2(Quary a, Quary b) { return a.x < b.x; }

void get()
{
    int x = lower_bound(A, A + sz, q[pq].y) - A; x++;
    if (sum(x) > ) ans[q[pq].id] = ; else ans[q[pq].id] = ; pq++;
}

void insert()
{
    int L = lower_bound(A, A + sz, s[ps].y1) - A; L++;
    int R = lower_bound(A, A + sz, s[ps].y2) - A; R++;
    update(L, R, s[ps].f); ps++;
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &nq);
        for (int i = ; i <= n; i++) scanf("%d", &v[i]);

        ns = ; sz = ;
        for (int i = ; i <= n; i++)
        {
            int l = , r = , L = , R = , w = , b = ;
            if (i & ) AddSeg(, v[i], , ); else AddSeg(, , , v[i]);
            for (int j = i + ; j <= n; j++)
            {
                l = r = w; L = R = b;
                if (i & ) r = r + v[i]; else R = R + v[i];
                if (j & ) r = r + v[j]; else R = R + v[j];
                AddSeg(l, r, L, R);
                if (j & ) w = w + v[j]; else b = b + v[j];
            }
        }

        for (int i = ; i < nq; i++)
        {
            scanf("%d%d", &q[i].x, &q[i].y); q[i].id = i;
            A[sz++] = q[i].y;
        }

        sort(s, s + ns, cmp); sort(q, q + nq, cmp2);
        sort(A, A + sz); sz = unique(A, A + sz) - A;

        ps = , pq = ; memset(c, , sizeof c);
        while (pq < nq)
        {
            if (ps == ns) get();
            else
            {
                if (s[ps].x < q[pq].x) insert();
                else if (s[ps].x > q[pq].x) get();
                else { if (s[ps].f == ) insert(); else get(); }
            }
        }
        for (int i = ; i < nq; i++) printf("%d", ans[i]); printf("\n");
    }
    return ;
}