标签:pre add put temp 问题 lower for lan cond
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] = Line{lx, ly, ry, 1}; L[++L_cnt] = Line{rx + 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] = Qus{x, 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; } /* */
HDU - 5741 Helter Skelter 扫描线 + 树状数组
标签:pre add put temp 问题 lower for lan cond
原文地址:https://www.cnblogs.com/CJLHY/p/11517157.html