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BZOJ3262 陌上花开
前人之述备矣、、、
树套树即BIT套treap 和 CQD分治 + BIT的方法都有了
于是就做好了233
1 /************************************************************** 2 Problem: 3262 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1356 ms 7 Memory:5888 kb 8 ****************************************************************/ 9 10 #include <cstdio>11 #include <algorithm>12 13 #define lowbit(x) x & -x14 using namespace std;15 const int N = 100005;16 17 struct data {18 int a, b, c, s, ans;19 }a[N], p[N];20 inline bool sort1_cmp (const data &a, const data &b) {21 return a.a == b.a ? (a.b == b.b ? a.c < b.c : a.b < b.b) : a.a < b.a;22 }23 inline bool operator < (const data &a, const data &b) {24 return a.b == b.b ? a.c < b.c : a.b < b.b;25 }26 inline bool operator == (const data &a, const data &b) {27 return a.a == b.a && a.b == b.b && a.c == b.c;28 }29 inline bool operator != (const data &a, const data &b) {30 return !(a == b);31 }32 33 int tot, n, m, BIT[N << 1], ans[N];34 35 inline int read() {36 int x = 0;37 char ch = getchar();38 while (ch < ‘0‘ || ‘9‘ < ch)39 ch = getchar();40 while (‘0‘ <= ch && ch <= ‘9‘) {41 x = x * 10 + ch - ‘0‘;42 ch = getchar();43 }44 return x;45 }46 47 inline void update(int x, int del) {48 while (x <= m)49 BIT[x] += del, x += lowbit(x);50 }51 52 inline int query(int x) {53 int res = 0;54 while (x)55 res += BIT[x], x -= lowbit(x);56 return res;57 }58 59 void work(int l, int r) {60 if (l == r) return;61 int mid = l + r >> 1, i, j;62 work(l, mid), work(mid + 1, r);63 sort(p + l, p + mid + 1), sort(p + mid + 1, p + r + 1);64 for (i = l, j = mid + 1; j <= r; ++j) {65 for (; i <= mid && p[i].b <= p[j].b; ++i)66 update(p[i].c, p[i].s);67 p[j].ans += query(p[j].c);68 }69 for (j = l; j < i; ++j)70 update(p[j].c, -p[j].s);71 }72 73 int main() {74 int i, cnt;75 tot = read(), m = read();76 for (i = 1; i <= tot; ++i)77 a[i].a = read(), a[i].b = read(), a[i].c = read();78 sort(a + 1, a + tot + 1, sort1_cmp);79 for (cnt = 1, i = 1; i <= tot; ++i, ++cnt)80 if (a[i] != a[i + 1]) {81 p[++n] = a[i];82 p[n].s = cnt;83 cnt = 0;84 }85 work(1, n);86 for (i = 1; i <= n; ++i)87 ans[p[i].ans + p[i].s - 1] += p[i].s;88 for (i = 0; i < tot; ++i)89 printf("%d\n", ans[i]);90 return 0;91 }
BZOJ3262 陌上花开
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