1. 1077 : RMQ问题再临-线段树
http://hihocoder.com/problemset/problem/1077
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<string.h> #define N 10005 #define min(a,b) (a<b?a:b) int n; struct node { int l, r; int min; }node[N << 2]; int BuildTree(int l, int r, int k) { // printf("%d %d %d ", l, r, k); node[k].l = l; node[k].r = r; if (l == r) { scanf("%d", &node[k].min); return 0; } BuildTree(l, (l + r) / 2, k * 2); BuildTree((l + r) / 2 + 1, r, k * 2 + 1); node[k].min = min(node[k * 2].min, node[k * 2 + 1].min); return 0; } int Query(int l, int r, int k) { if (l <= node[k].l && r >= node[k].r) return node[k].min; int minl = N, minr = N; if (l <= node[k * 2].r) minl = Query(l, r, k * 2); if (r >= node[k * 2 + 1].l) minr = Query(l, r, k * 2 + 1); return min(minl,minr); } int Change(int p, int w, int k) { if (node[k].l == node[k].r) { node[k].min = w; return 0; } if (p <= node[k * 2].r) Change(p, w, k * 2); if (p >= node[k * 2 + 1].l) Change(p, w, k * 2 + 1); node[k].min = min(node[k * 2].min, node[k * 2 + 1].min); return 0; } int main() { int q, k, a, b, i; scanf("%d", &n); BuildTree(1,n,1); scanf("%d", &q); for (i = 1; i <= q; i++) { scanf("%d%d%d", &k, &a, &b); k ? Change(a, b, 1) : printf("%d ",Query(a, b, 1)); } return 0; }
2. #1078 : 线段树的区间修改（LazyTag）
http://hihocoder.com/problemset/problem/1078
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #define N 100005 struct node { int l, r; int len; int sum; int lazytag; //为0时表示没有lazytag，非0时表示要修改的单价 }node[N<<2]; void BuildTree(int l, int r, int k) { node[k].l = l; node[k].r = r; node[k].len = r - l + 1; node[k].lazytag = 0; if (l == r) { scanf("%d", &node[k].sum); return; } BuildTree(l, (l + r) / 2, k * 2); BuildTree((l + r) / 2 + 1, r, k * 2 + 1); node[k].sum = node[k * 2].sum + node[k * 2 + 1].sum; } int Query(int l, int r, int k) { if (node[k].l >= l && node[k].r <= r) return node[k].sum; if (node[k].lazytag) { node[k * 2].lazytag = node[k].lazytag; node[k * 2].sum = node[k].lazytag * node[k * 2].len; node[k * 2 + 1].lazytag = node[k].lazytag; node[k * 2 + 1].sum = node[k].sum - node[k * 2].sum; node[k].lazytag = 0; } int mid = (node[k].l+node[k].r)/2, lsum = 0, rsum = 0; if (l <= mid) lsum = Query(l, r, k * 2); if (r > mid) rsum = Query(l, r, k * 2 + 1); return lsum + rsum; } void Change(int l, int r, int p, int k) { // printf("%d %d %d %d %d ", l, r, k,node[k].l,node[k].r); if (l == node[k].l && r == node[k].r) { node[k].lazytag = p; node[k].sum = p * node[k].len; return; } if (node[k].lazytag) { node[k * 2].lazytag = node[k].lazytag; node[k * 2].sum = node[k].lazytag * node[k * 2].len; node[k * 2 + 1].lazytag = node[k].lazytag; node[k * 2 + 1].sum = node[k].sum - node[k * 2].sum; node[k].lazytag = 0; } int mid = (node[k].l + node[k].r) / 2; if (l <= mid) Change(l, r<mid?r:mid, p, k * 2); if (r > mid) Change(l>(mid+1)?l:(mid+1), r, p, k * 2 + 1); node[k].sum = node[k * 2].sum + node[k * 2 + 1].sum; } void printTree() { printf("********** "); for (int i = 1; i <= 25; i++) { printf("(%d,%d) %d ", node[i].l, node[i].r, node[i].sum); } printf("********** "); } int main() { int n, q, k, l, r, p, i; scanf("%d", &n); BuildTree(1, n, 1); printTree(); scanf("%d", &q); for (i = 0; i < q; i++) { scanf("%d%d%d", &k, &l, &r); if (0 == k) printf("%d ", Query(l, r, 1)); else { scanf("%d", &p); Change(l, r, p, 1); } // printTree(); } return 0; }
3. #1080 : 更为复杂的买卖房屋姿势（两种LazyTag）
http://hihocoder.com/problemset/problem/1080
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #define N 100005 struct node { int l, r; int len; int sum; int lazyTagAdd, lazyTagSet; }node[N<<2]; void BuildTree(int l, int r, int k) { node[k].l = l; node[k].r = r; node[k].len = r - l + 1; node[k].lazyTagAdd = 0; node[k].lazyTagSet = 0; if (l == r) { scanf("%d", &node[k].sum); return; } int mid = (l + r) >> 1; BuildTree(l, mid, k << 1); BuildTree(mid + 1, r, (k << 1) | 1); node[k].sum = node[k << 1].sum + node[k << 1 | 1].sum; } void Change(int op, int l, int r, int v, int k) { // printf("%d %d %d %d %d ", op, l, r, v, k); if (node[k].len > 1) { if (node[k].lazyTagSet) { node[k << 1].lazyTagSet = node[k].lazyTagSet; node[k << 1].sum = node[k].lazyTagSet * node[k << 1].len; node[k << 1].lazyTagAdd = 0; node[k << 1 | 1].lazyTagSet = node[k].lazyTagSet; node[k << 1 | 1].sum = node[k].lazyTagSet * node[k << 1 | 1].len; node[k << 1 | 1].lazyTagAdd = 0; node[k].lazyTagSet = 0; } if (node[k].lazyTagAdd) { node[k << 1].lazyTagAdd += node[k].lazyTagAdd; node[k << 1].sum += node[k].lazyTagAdd * node[k << 1].len; node[k << 1 | 1].lazyTagAdd += node[k].lazyTagAdd; node[k << 1 | 1].sum += node[k].lazyTagAdd * node[k << 1 | 1].len; node[k].lazyTagAdd = 0; } } if (l <= node[k].l && r >= node[k].r) { if (0 == op) //add { node[k].lazyTagAdd = v; node[k].sum += v* node[k].len; } else { node[k].lazyTagSet = v; node[k].sum = v * node[k].len; } return; } int mid = (node[k].l + node[k].r) >> 1; if (l <= mid) Change(op, l, r, v, k << 1); if (r > mid) Change(op, l, r, v, k << 1 | 1); node[k].sum = node[k << 1].sum + node[k << 1 | 1].sum; } void PrintTree() { printf(" ****** "); for (int i = 1; i < 30; i++) { printf("%d: %d,%d %d ", i, node[i].l, node[i].r, node[i].sum); } printf(" ****** "); } int main() { int n, m, op, l, r, v, i; scanf("%d%d", &n, &m); BuildTree(0, n, 1); // for (i = 1; i < 30; i++) // printf("%d: %d,%d %d ", i, node[i].l, node[i].r, node[i].sum); PrintTree(); for (i = 0; i < m; i++) { scanf("%d%d%d%d", &op, &l, &r, &v); Change(op, l, r, v, 1); printf("%d ", node[1].sum); PrintTree(); } return 0; }
4. #1079 : 离散化
http://hihocoder.com/problemset/problem/1079
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<cstring> #include<algorithm> using namespace std; #define N 100005 int s[N], e[N] ,p[N<<2]; bool visit[N]; typedef struct Poster { int v; //区间端点 int k; //序号 }Poster; Poster P[N << 1]; bool cmp(Poster a, Poster b) { return a.v < b.v; } void BuildTree(int k, int l, int r, int num) { if (s[num] <= l && e[num] >= r) { p[k] = num; //结点k存放的是第num张海报 return; } if (p[k]) { p[k << 1] = p[k]; p[k << 1 | 1] = p[k]; p[k] = 0; } int mid = (l + r) / 2; if (s[num] < mid) BuildTree(k << 1, l, mid, num); if (e[num] > mid) BuildTree(k << 1 | 1, mid, r, num); } void Check(int k, int l, int r) { if (l + 1 == r) { visit[p[k]] = true; return; } if (p[k]) { p[k << 1] = p[k]; p[k << 1 | 1] = p[k]; p[k] = 0; } int mid = (l + r) / 2; Check(k << 1, l, mid); Check(k << 1 | 1, mid, r); } int main() { int n, l, i, j; cin >> n >> l; for (i = 1; i <= n; i++) { cin >> P[(i << 1) - 1].v; P[(i<<1)-1].k = i; cin >> P[i << 1].v; P[i<<1].k = i + n; } sort(P + 1, P + (n << 1) + 1, cmp); P[(n << 1) + 1].v = -1; for (i = 1, j = 1; i <= (n << 1); i++) { if (P[i].k <= n) s[P[i].k] = j; else e[P[i].k - n] = j; if (P[i].v != P[i + 1].v) j++; } memset(p, 0, sizeof(p)); for (i = 1; i <= n; i++) BuildTree(1, 1, j, i); memset(visit, 0, sizeof(visit)); Check(1, 1, j); for (i = 1, j = 0; i <= n; i++) { if (visit[i]) j++; } cout << j << endl; return 0; }
5. 1116 : 计算
http://hihocoder.com/problemset/problem/1116?sid=769706
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<string.h> const int MOD = 10007; const int N = 100005; typedef struct interval { //int left, right; int sum; int pre; //区间内以left为前缀的所有乘积的总和 int suf; //suffix, 区间内以right为后缀的所有乘积的总和 int prd; //product, 区间内所有数的乘积，为求父节点的pre }interval; interval in[N << 1]; void Update(int k) { while (k) { in[k].pre = (in[k << 1].pre + in[k << 1].prd * in[k << 1 | 1].pre) % MOD; in[k].suf = (in[k << 1].suf * in[k << 1 | 1].prd + in[k << 1 | 1].suf) % MOD; in[k].prd = (in[k << 1].prd * in[k << 1 | 1].prd) % MOD; in[k].sum = (in[k << 1].sum + in[k << 1 | 1].sum + in[k << 1].suf * in[k << 1 | 1].pre) % MOD; k = k >> 1; } } void Change(int k, int left, int right, int i, int x) { if (left == right) { in[k].sum = x % MOD; in[k].pre = x % MOD; in[k].suf = x % MOD; in[k].prd = x % MOD; Update(k >> 1); return; } int mid = (left + right) / 2; if (i <= mid) Change(k << 1, left, mid, i, x); else Change(k << 1 | 1, mid + 1, right, i, x); } int main() { int n, q, x, i; scanf("%d%d", &n, &q); memset(in, 0, sizeof(in)); while (q--) { scanf("%d%d", &i, &x); Change(1, 1, n, i, x); printf("%d ", in[1].sum); } return 0; }