数据结构（荣誉）实验六 线段树
作者：互联网
目录
1.敌兵布阵
题目描述
C国的死对头A国这段时间正在进行军事演习，所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地，Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段，所以每个工兵营地的人数C国都掌握的一清二楚，每个工兵营地的人数都有可能发生变动，可能增加或减少若干人手，但这些都逃不过C国的监视。
输入
第一行一个整数T，表示有T组数据。
每组数据第一行有一个正整数N（N<=50000），表示敌人有N个工兵营地，接下来一行有N个正整数，第i个正整数ai代表第i个工兵营地里开始时有ai个人（1<=ai<=50）。
接下来每行有一条命令，命令有4种形式：
(1) Add i j，i和j为正整数，表示第i个营地增加j个人（j不超过30）；
(2) Sub i j，i和j为正整数，表示第i个营地减少j个人（j不超过30）；
(3) Query i j，i和j为正整数，i<=j，表示询问第i到第j个营地的总人数；
(4) End 表示结束，这条命令在每组数据最后出现。
输出
对第i组数据，首先输出“Case i:”和回车，
对于每个Query询问，输出一个整数并回车，表示询问的段中的总人数，这个数保持在int以内。
样例输入
1
10
1 2 3 4 5 6 7 8 9 10
Query 1 3
Add 3 6
Query 2 7
Sub 10 2
Add 6 3
Query 3 10
End
样例输出
Case 1:
6
33
59
题解
#include <algorithm>
#include <iostream>
using namespace std;
#define MID(l, r) (l + r) >> 1
#define lson root << 1
#define rson root << 1 | 1
const int maxn = 5e4 + 5;
struct Tree
{
int l, r;
int sum;
} tree[maxn * 3];
int arr[maxn], ans;
void Build(int root, int l, int r)
{
tree[root].l = l;
tree[root].r = r;
if (l == r)
tree[root].sum = arr[l];
else
{
int mid = MID(l, r);
Build(lson, l, mid);
Build(rson, mid + 1, r);
tree[root].sum = tree[lson].sum + tree[rson].sum;
}
}
void Query(int root, int x, int y)
{
if (x <= tree[root].l && y >= tree[root].r)
ans += tree[root].sum;
else
{
int mid = MID(tree[root].l, tree[root].r);
if (x > mid)
{
Query(rson, x, y);
}
else if (y <= mid)
{
Query(lson, x, y);
}
else
{
Query(lson, x, y);
Query(rson, x, y);
}
}
}
void Update(int root, int x, int y)
{
tree[root].sum += y;
if (tree[root].l == x && tree[root].r == x)
return;
int mid = MID(tree[root].l, tree[root].r);
if (x > mid)
Update(rson, x, y);
else
Update(lson, x, y);
}
int main()
{
int T;
cin >> T;
for (int flag = 1; flag <= T; flag++)
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> arr[i];
Build(1, 1, n);
cout << "Case " << flag << ":" << endl;
string ope;
while (true)
{
cin >> ope;
if (ope == "End")
{
break;
}
if (ope == "Add")
{
int x, y;
cin >> x >> y;
Update(1, x, y);
continue;
}
if (ope == "Sub")
{
int x, y;
cin >> x >> y;
Update(1, x, -y);
}
if (ope == "Query")
{
int x, y;
cin >> x >> y;
ans = 0;
Query(1, x, y);
cout << ans << endl;
}
}
}
return 0;
}
2. 最大数
题目描述
输入
输出
样例输入
5 100
A 96
Q 1
A 97
Q 1
Q 2
样例输出
96
93
96
题解
#include <iostream>
#include <cstdio>
#include <algorithm>
#define MAX 100000
using namespace std;
int mx[MAX * 4], a[MAX], m, d, x, n, now;
void Build(int k, int l, int r, int x) {
if (l == r && l == x) {
mx[k] = a[l];
return;
}
int mid = (l + r) >> 1;
if (x <= mid) {
Build(k << 1, l, mid, x);
} else {
Build(k << 1 | 1, mid + 1, r, x);
}
mx[k] = max(mx[k << 1], mx[k << 1 | 1]);
}
int query(int k, int l, int r, int x, int y) {
if (l == x && r == y) {
return mx[k];
}
int mid = (l + r) >> 1;
if (y <= mid) {
return query(k << 1, l, mid, x, y);
}
if (x > mid) {
return query(k << 1 | 1, mid + 1, r, x, y);
}
return
max(query(k << 1, l, mid, x, mid), query(k << 1 | 1, mid + 1, r, mid + 1, y));
}
int main() {
cin >> m >> d;
for (int i = 1; i <= m; i++) {
char op;
cin >> op;
if (op == 'A') {
cin >> x;
x += now;
x %= d;
a[++n] = x;
Build(1, 1, m, n);
} else {
cin >> x;
now = query(1, 1, m, n - x + 1, n);
printf("%d\n", now);
}
}
}
3.Distinct Characters Queries(英文题面)(线段树)
题目描述
输入
输出
样例输入
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
样例输出
3
1
2
题解
#include <iostream>
#include <set>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e5 + 1;
set<int> st[99];
char c[N];
int main() {
cin >> (c + 1);
int n;
cin >> n;
int len = strlen(c + 1);
for (int i = 1; i <= len; ++i) {
st[c[i] - 'a'].insert(i);
}
int f, t, d;
char k;
for (int i = 1; i <= n; ++i) {
cin >> f;
if (f == 1) {
cin >> t >> k;
st[c[t] - 'a'].erase(t);
st[k - 'a'].insert(t);
c[t] = k;
} else {
cin >> t >> d;
int sum = 0;
for (int j = 0; j < 26; ++j) {
auto p = st[j].lower_bound(t);
if (p != st[j].end() && *p <= d) sum++;
}
cout << sum << endl;
}
}
}
4.Kth number
题目描述
Give you a sequence and ask you the kth big number of a inteval.
输入
The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
输出
For each test case, output m lines. Each line contains the kth big number.
样例输入
1
10 1
1 4 2 3 5 6 7 8 9 0
1 3 2
样例输出
2
题解
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn = 1e5 + 7;
int t, n, m, cnt, root[maxn], a[maxn], x, y, k;
struct node {
int l, r, sum;
} T[maxn * 25];
vector<int> v;
int getid(int x) {
return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
void update(int l, int r, int &x, int y, int pos) {
T[++cnt] = T[y], T[cnt].sum++, x = cnt;
if (l == r) {
return;
}
int mid = (l + r) / 2;
if (mid >= pos) {
update(l, mid, T[x].l, T[y].l, pos);
} else {
update(mid + 1, r, T[x].r, T[y].r, pos);
}
}
int query(int l, int r, int x, int y, int k) {
if (l == r) {
return l;
}
int mid = (l + r) / 2;
int sum = T[T[y].l].sum - T[T[x].l].sum;
if (sum >= k) {
return query(l, mid, T[x].l, T[y].l, k);
} else {
return query(mid + 1, r, T[x].r, T[y].r, k - sum);
}
}
int main() {
scanf("%d", &t);
while (t--) {
v.clear();
cnt = 0;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
v.push_back(a[i]);
}
sort(v.begin(), v.end());
v.erase(unique(v.begin(), v.end()), v.end());
for (int i = 1; i <= n; i++) {
update(1, n, root[i], root[i - 1], getid(a[i]));
}
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &x, &y, &k);
printf("%d\n", v[query(1, n, root[x - 1], root[y], k) - 1]);
}
}
return 0;
}
标签：荣誉,int,tree,线段,样例,mid,数据结构,root,sum 来源： https://blog.csdn.net/m0_46326495/article/details/117927399