HDU-1698-Just a Hook
链接:https://vjudge.net/problem/HDU-1698#author=0
题意:
给n个数,默认为1,q次操作,将x-y内的数改成1-3中的一种。
求n个数的和。
思路:
线段树,区间更新。
代码:
#include <iostream> #include <memory.h> #include <vector> #include <map> #include <algorithm> using namespace std; typedef long long LL; const int MAXN = 1e5 + 10; int segments[MAXN * 4]; int add[MAXN * 4]; void Push_down(int root, int l, int r) { if (add[root]) { add[root << 1] = add[root]; add[root << 1 | 1] = add[root]; int mid = (l + r) / 2; segments[root << 1] = add[root] * (mid - l + 1); segments[root << 1 | 1] = add[root] * (r - mid); add[root] = 0; } } void Build_tree(int root, int l, int r) { if (l == r) { segments[root] = 1; return ; } int mid = (l + r) / 2; Build_tree(root << 1, l, mid); Build_tree(root << 1 | 1, mid + 1, r); segments[root] = segments[root << 1] + segments[root << 1 | 1]; } void Update_tree(int root, int l, int r, int ql, int qr, int c) { if (ql > r || qr < l) return ; if (ql <= l && r <= qr) { segments[root] = (r - l + 1) * c; add[root] = c; } else { Push_down(root, l, r); int mid = (l + r) / 2; Update_tree(root << 1, l, mid, ql, qr, c); Update_tree(root << 1 | 1, mid + 1, r, ql, qr, c); segments[root] = segments[root << 1] + segments[root << 1 | 1]; } } int Query(int root, int l, int r, int w) { if (l == r) return segments[root]; int mid = (l + r) / 2; if (w <= mid) return Query(root << 1, l, mid, w); else return Query(root << 1 | 1, mid + 1, r, w); } int main() { int t, cnt = 0; scanf("%d", &t); while (t--) { int n, q; int l, r, op; scanf("%d%d", &n, &q); Build_tree(1, 1, n); memset(add, 0, sizeof(add)); for (int i = 1;i <= q;i++) { scanf("%d%d%d", &l, &r, &op); Update_tree(1, 1, n, l, r, op); } printf("Case %d: The total value of the hook is %d.\n", ++cnt, segments[1]); } return 0; }