HDU - 5741 Helter Skelter 扫描线 + 树状数组

Posted 2022-10-26 cjlhy

HDU - 5741

我们枚举段的起点和终点， 那么每一种情况0的范围是[lx, rx]， 1的出现范围是[ly, ry]， 可以在二维平面上用矩形表示。

然后问题就变成了询问点有没有被至少一个矩形覆盖， 扫描线 + 树状数组就可以了。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 3e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) a += b; if(a >= mod) a -= mod;
template<class T, class S> inline void sub(T& a, S b) a -= b; if(a < 0) a += mod;
template<class T, class S> inline bool chkmax(T& a, S b) return a < b ? a = b, true : false;
template<class T, class S> inline bool chkmin(T& a, S b) return a > b ? a = b, true : false;

struct Line 
    LL x, y1, y2;
    bool operator < (const Line &rhs) const 
        return x < rhs.x;
    
    int w;
;

struct Qus 
    bool operator < (const Qus &rhs) const 
        return x < rhs.x;
    
    LL x, y, id;
;

LL hs[N], hs_cnt;

struct Bit 
    int a[N];
    void init() 
        for(int i = 1; i <= hs_cnt; i++) 
            a[i] = 0;
        
    
    inline void modify(int x, int v) 
        for(int i = x; i <= hs_cnt; i += i & -i) 
            a[i] += v;
        
    
    inline int sum(int x) 
        int ans = 0;
        for(int i = x; i; i -= i & -i) 
            ans += a[i];
        
        return ans;
    
;

int n, m, mx[2], a[N], ans[N];
LL sum[2][N];

int L_cnt;
Line L[N];

int Q_cnt;
Qus Q[N];

Bit bit;

void init() 
    mx[0] = mx[1] = 0;
    L_cnt = Q_cnt = hs_cnt = 0;


int main() 
    int T;
    scanf("%d", &T);
    while(T--) 
        scanf("%d%d", &n, &m);
        init();
        for(int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
            sum[0][i] = sum[0][i - 1];
            sum[1][i] = sum[1][i - 1];
            sum[!(i & 1)][i] += a[i];
            chkmax(mx[!(i & 1)], a[i]);
        
        for(int i = 1; i <= n; i++) 
            for(int j = i + 1; j <= n; j++) 
                LL lx = 0, ly = 0, rx = 0, ry = 0;
                if(i + 1 < j) 
                    lx = sum[0][j - 1] - sum[0][i];
                    ly = sum[1][j - 1] - sum[1][i];
                
                rx = lx; ry = ly;
                if(i & 1) rx += a[i];
                else ry += a[i];
                if(j & 1) rx += a[j];
                else ry += a[j];
                L[++L_cnt] = Linelx, ly, ry, 1;
                L[++L_cnt] = Linerx + 1, ly, ry, -1;
                hs[++hs_cnt] = ly;
                hs[++hs_cnt] = ry;
            
        
        for(int i = 1; i <= m; i++) 
            LL x, y; scanf("%lld%lld", &x, &y);
            if(!x) 
                ans[i] = mx[1] >= y;
            
            else if(!y) 
                ans[i] = mx[0] >= x;
            
            else 
                hs[++hs_cnt] = y;
                Q[++Q_cnt] = Qusx, y, i;
            
        
        sort(hs + 1, hs + 1 + hs_cnt);
        hs_cnt = unique(hs + 1, hs + 1 + hs_cnt) - hs - 1;
        bit.init();
        for(int i = 1; i <= L_cnt; i++) 
            L[i].y1 = lower_bound(hs + 1, hs + 1 + hs_cnt, L[i].y1) - hs;
            L[i].y2 = lower_bound(hs + 1, hs + 1 + hs_cnt, L[i].y2) - hs;
        
        for(int i = 1; i <= Q_cnt; i++) 
            Q[i].y = lower_bound(hs + 1, hs + 1 + hs_cnt, Q[i].y) - hs;
        
        sort(L + 1, L + 1 + L_cnt);
        sort(Q + 1, Q + 1 + Q_cnt);
        for(int i = 1, j = 1; i <= Q_cnt; i++) 
            while(j <= L_cnt && L[j].x <= Q[i].x) 
                bit.modify(L[j].y1, L[j].w);
                bit.modify(L[j].y2 + 1, -L[j].w);
                j++;
            
            ans[Q[i].id] = bit.sum(Q[i].y);
        
        for(int i = 1;  i <= m; i++) 
            putchar(ans[i] ? ‘1‘ : ‘0‘);
        
        puts("");
    
    return 0;


/*
*/