题目大意：两种操作，1是改动区间上l到r上面德值为x，2是询问l到r区间总的改动值。
解题思路：线段树模板题。
#include#include #include #include using namespace std;const int maxn = 5*1e5;typedef long long ll;ll s[maxn], del[maxn], mark[maxn];void build (int u, int l, int r) { if (l < r) { int mid = (l + r)/2; build(u*2, l, mid); build(u*2+1, mid+1, r); } else mark[u] = l;}ll query (int u, int l, int r, int x, int y) { if (y < l || x > r) return 0; if (x <= l && r <= y) return s[u]; int mid = (l + r) / 2; ll L = query(u*2, l, mid, x, y); ll R = query(u*2+1, mid+1, r, x, y); return L + R + max(0, min(r, y) - max(l, x) + 1) * del[u];}void clear (int u, int l, int r, int z) { if (mark[u]) { del[u] += abs(mark[u]-z); s[u] += 1LL * (r-l+1) * abs(mark[u]-z); mark[u] = 0; } else { if (l == r) return; int mid = (l + r) / 2, lc = u*2, rc = u*2+1; clear(lc, l, mid, z); clear(rc, mid+1, r, z); s[u] = s[lc] + s[rc] + 1LL * (r-l+1) * del[u]; }}void modify (int u, int l, int r, int x, int y, int z) { // printf("%d %d %d %d %d %d\n", u, l, r, x, y, z); if (y < l || x > r) return; if (x <= l && r <= y) { clear(u, l, r, z); mark[u] = z; return; } int mid = (l + r) / 2, lc = u*2, rc = u*2+1; if (mark[u]) { mark[lc] = mark[rc] = mark[u]; mark[u] = 0; } modify(lc, l, mid, x, y, z); modify(rc, mid+1, r, x, y, z); s[u] = s[lc] + s[rc] + 1LL * (r-l+1) * del[u];}int main () { int n, m; scanf("%d%d", &n, &m); /* memset(sum, 0, sizeof(sum)); memset(del, 0, sizeof(del)); memset(mark, 0, sizeof(mark)); */ build(1, 1, n); int type, x, y, z;; for (int i = 0; i < m; i++) { scanf("%d%d%d", &type, &x, &y); if (type == 1) { scanf("%d", &z); modify(1, 1, n, x, y, z); } else printf("%lld\n", query(1, 1, n, x, y)); } return 0;}