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题目链接:http://codeforces.com/problemset/problem/295/A
我的做法,两次线段树
#include <cstdio> #include <cstring> const int N = 100005; long long sumv[N * 3]; long long add[N * 3]; long long a[N]; struct opera { int l, r; long long d; } op[N]; void pushdown(int o, int l, int r) { int m = (l + r) >> 1; sumv[o << 1] += add[o] * (m - l + 1); add[o << 1] += add[o]; sumv[o << 1 | 1] += add[o] * (r - m); add[o << 1 | 1] += add[o]; add[o] = 0; } int ql, qr; long long val; void update(int o, int l, int r) { if (ql <= l && qr >= r) { sumv[o] += (r - l + 1) * val; add[o] += val; return ; } if (add[o]) pushdown(o, l, r); int m = (l + r) >> 1; if (m < qr) update(o << 1 | 1, m + 1, r); if (m >= ql) update(o << 1, l, m); sumv[o] = sumv[o << 1] + sumv[o << 1 | 1]; } int q; long long query(int o, int l, int r) { if (l == r) return sumv[o]; if (add[o]) pushdown(o, l, r); int m = (l + r) >> 1; if (m < q) return query(o << 1 | 1, m + 1, r); else return query(o << 1, l, m); } int main() { //freopen("a.in", "r", stdin); int n, m, k; int i; scanf("%d%d%d", &n, &m, &k); for (i = 1; i <= n; ++i) scanf("%lld", a + i); for (i = 1; i <= m; ++i) { scanf("%d%d%lld", &op[i].l, &op[i].r, &op[i].d); } val = 1; for (i = 1; i <= k; ++i) { scanf("%d%d", &ql, &qr); update(1, 1, m); } for (i = 1; i <= m; ++i) { q = i; op[i].d *= query(1, 1, m); } memset(sumv, 0, sizeof sumv); memset(add, 0, sizeof add); for (i = 1; i <= m; ++i) { ql = op[i].l; qr = op[i].r; val = op[i].d; update(1, 1, n); } for (i = 1; i <= n; ++i) { q = i; printf("%lld", a[i] + query(1, 1, n)); if (i != n) printf(" "); } return 0; }
后来看了学长的代码,又写了一遍.......:
#include <cstdio> const int N = 100005; long long l[N]; long long r[N]; long long d[N]; long long a[N]; long long op[N]; long long b[N]; main() { //freopen("in.txt", "r", stdin); long long n, m, k; scanf("%lld%lld%lld", &n, &m, &k); for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]); for (int i = 1; i <= m; ++i) scanf("%lld%lld%lld", &l[i], &r[i], &d[i]); long long x, y; for (int i = 1; i <= k; ++i) { scanf("%lld%lld", &x, &y); ++op[x]; --op[y+1]; } for (int i = 1; i <= m; ++i) { op[i] += op[i - 1]; d[i] *= op[i]; } for (int i = 1; i <= m; ++i) { b[l[i]] += d[i]; b[r[i]+1] -= d[i]; } for (int i = 1; i <= n; ++i) { b[i] += b[i - 1]; a[i] += b[i]; } printf("%lld", a[1]); for (int i = 2; i <= n; ++i) printf(" %lld", a[i]); }
CodeForces Round #179 (295A) - Greg and Array
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原文地址:http://www.cnblogs.com/wenruo/p/4692999.html