wugensheng

3061

线段树

1. 两种操作合并成一种
2. 合并后的操作需要注意蓝标记向下传递的时候，的更新的情况
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef long long ll;
int n, m, p;
int a[N];
struct node {
int l, r;
} tr[N * 4];

void eval(node &root, int mul, int add) {
root.sum = (root.sum * mul % p + (ll)(root.r - root.l + 1) * add) % p;
root.mul = root.mul * mul % p;
}

void pushup(int u) {
tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % p;
}

void pushdown(int u) {
eval(tr[u << 1 | 1], tr[u].mul, tr[u].add);
tr[u].mul = 1;
}

void build (int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l] % p, 1, 0};
else {
tr[u] = {l, r, 0, 1, 0};  // 初始化，太坑了
int mid = (l + r) >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int l, int r, int mul, int add) {
if (tr[u].l >= l && tr[u].r <= r) {
} else {
pushdown(u);
int mid = (tr[u].l + tr[u].r) >> 1;
if (l <= mid) modify(u << 1, l, r, mul, add);
if (r > mid) modify(u << 1 | 1, l, r, mul, add);
pushup(u);
}
}

ll query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
pushdown(u);
int mid = (tr[u].l + tr[u].r) >> 1;
ll res = 0;
if (l <= mid) res += query(u << 1, l, r) % p;
if (r > mid) res = (res + query(u << 1 | 1, l, r)) % p;
return res;
}

int main() {
cin >> n >> p;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
build(1, 1, n);
int k, t, g, c;
cin >> m;
while (m--) {
scanf("%d%d%d", &k, &t, &g);
if (k == 1 || k == 2) {
scanf("%d", &c);
if (k == 1) modify(1, t, g, c, 0);
else modify(1, t, g, 1, c);
} else if (k == 3) {
printf("%lld\n", query(1, t, g));
}
}
return 0;
}


线段树

1. 两种操作合并成一种
2. 合并后的操作需要注意蓝标记向下传递的时候，的更新的情况
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef long long ll;
int n, m, p;
int a[N];
struct node {
int l, r;
} tr[N * 4];

void eval(node &root, int mul, int add) {
root.sum = (root.sum * mul % p + (ll)(root.r - root.l + 1) * add) % p;
root.mul = root.mul * mul % p;
}

void pushup(int u) {
tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % p;
}

void pushdown(int u) {
eval(tr[u << 1 | 1], tr[u].mul, tr[u].add);
tr[u].mul = 1;
}

void build (int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l] % p, 1, 0};
else {
tr[u] = {l, r, 0, 1, 0};  // 初始化，太坑了
int mid = (l + r) >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int l, int r, int mul, int add) {
if (tr[u].l >= l && tr[u].r <= r) {
} else {
pushdown(u);
int mid = (tr[u].l + tr[u].r) >> 1;
if (l <= mid) modify(u << 1, l, r, mul, add);
if (r > mid) modify(u << 1 | 1, l, r, mul, add);
pushup(u);
}
}

ll query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
pushdown(u);
int mid = (tr[u].l + tr[u].r) >> 1;
ll res = 0;
if (l <= mid) res += query(u << 1, l, r) % p;
if (r > mid) res = (res + query(u << 1 | 1, l, r)) % p;
return res;
}

int main() {
cin >> n >> p;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
build(1, 1, n);
int k, t, g, c;
cin >> m;
while (m--) {
scanf("%d%d%d", &k, &t, &g);
if (k == 1 || k == 2) {
scanf("%d", &c);
if (k == 1) modify(1, t, g, c, 0);
else modify(1, t, g, 1, c);
} else if (k == 3) {
printf("%lld\n", query(1, t, g));
}
}
return 0;
}


字典树

1. 区间修改用懒标记
2. 在每次分裂之前，都需要进行pushdown

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef long long ll;
int a[N];
int n, m;
struct node {
int l, r;
} tr[N * 4];

void pushup(int u) {
tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u) {
auto &l = tr[u << 1], &r = tr[u << 1 | 1];
}

void build(int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l], 0};
else {
tr[u] = {l, r};
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int l, int r, int d) {
if (tr[u].l >= l && tr[u].r <= r) tr[u].add += d, tr[u].sum += (ll)(tr[u].r - tr[u].l + 1) * d;
else {
pushdown(u);  //这里不可以省略，每次分裂之前都要向下传递
int mid = tr[u].l + tr[u].r >> 1;
if (l <= mid) modify(u << 1, l, r, d);
if (r > mid) modify(u << 1 | 1, l, r, d);
pushup(u);
}
}

ll query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
else {
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
ll sum = 0;
if (l <= mid) sum += query(u << 1, l, r);
if (r > mid) sum += query(u << 1 | 1, l, r);
//pushup(u);
return sum;
}
}

int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

build(1, 1, n);

char s[2];
int l, r, d;
while (m--) {
scanf("%s%d%d", s, &l, &r);
if (*s == 'C') {
scanf("%d", &d);
modify(1, l, r, d);
} else {
printf("%lld\n", query(1, l, r));
}
}

return 0;
}


字典树

1. 区间修改用懒标记
2. 在每次分裂之前，都需要进行pushdown

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef long long ll;
int a[N];
int n, m;
struct node {
int l, r;
} tr[N * 4];

void pushup(int u) {
tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u) {
auto &l = tr[u << 1], &r = tr[u << 1 | 1];
}

void build(int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l], 0};
else {
tr[u] = {l, r};
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int l, int r, int d) {
if (tr[u].l >= l && tr[u].r <= r) tr[u].add += d, tr[u].sum += (ll)(tr[u].r - tr[u].l + 1) * d;
else {
pushdown(u);  //这里不可以省略，每次分裂之前都要向下传递
int mid = tr[u].l + tr[u].r >> 1;
if (l <= mid) modify(u << 1, l, r, d);
if (r > mid) modify(u << 1 | 1, l, r, d);
pushup(u);
}
}

ll query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
else {
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
ll sum = 0;
if (l <= mid) sum += query(u << 1, l, r);
if (r > mid) sum += query(u << 1 | 1, l, r);
//pushup(u);
return sum;
}
}

int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

build(1, 1, n);

char s[2];
int l, r, d;
while (m--) {
scanf("%s%d%d", s, &l, &r);
if (*s == 'C') {
scanf("%d", &d);
modify(1, l, r, d);
} else {
printf("%lld\n", query(1, l, r));
}
}

return 0;
}


线段树

1. 根据题目分析，节点需要添加的新的信息，新信息如何维护，继续思考，直到全部信息可知
2. query的时候，需要分情况，因为查询的东西比较特殊，需要利用返回值

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 500010, M = 100010;
int a[N];
int n, m;
struct Node {
int l, r;
int sum, lmax, rmax, tmax;
} tr[N * 4];

void pushup(Node &u, Node &l, Node &r){
u.sum = l.sum + r.sum;
u.lmax = max(l.lmax, l.sum + r.lmax);
u.rmax = max(r.rmax, r.sum + l.rmax);
u.tmax = max(max(l.tmax, r.tmax), l.rmax + r.lmax);
}

void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l], a[l], a[l], a[l]};
else {
tr[u] = {l, r};
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int x, int v) {
if (tr[u].l == x && tr[u].r == x) tr[u] = {x, x, v, v, v, v};
else {
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
}

Node query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u];
else {
int mid = tr[u].l + tr[u].r >> 1;
if (r <= mid) return query(u << 1, l, r);
else if (l > mid) return query(u << 1 | 1, l, r);
else {
auto lnode = query(u << 1, l, r);
auto rnode = query(u << 1 | 1, l, r);
Node res;
pushup(res, lnode, rnode);
return res;
}
}
}

int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build(1, 1, n);

int x, y, k;
while (m--) {
scanf("%d%d%d", &k, &x, &y);
if (k == 1) {
if (x > y) swap(x, y);
auto t = query(1, x, y);
printf("%d\n", t.tmax);
} else {
modify(1, x, y);
}
}
return 0;
}


线段树

1. 根据题目分析，节点需要添加的新的信息，新信息如何维护，继续思考，直到全部信息可知
2. query的时候，需要分情况，因为查询的东西比较特殊，需要利用返回值

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 500010, M = 100010;
int a[N];
int n, m;
struct Node {
int l, r;
int sum, lmax, rmax, tmax;
} tr[N * 4];

void pushup(Node &u, Node &l, Node &r){
u.sum = l.sum + r.sum;
u.lmax = max(l.lmax, l.sum + r.lmax);
u.rmax = max(r.rmax, r.sum + l.rmax);
u.tmax = max(max(l.tmax, r.tmax), l.rmax + r.lmax);
}

void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r) {
if (l == r) tr[u] = {l, r, a[l], a[l], a[l], a[l]};
else {
tr[u] = {l, r};
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
}

void modify(int u, int x, int v) {
if (tr[u].l == x && tr[u].r == x) tr[u] = {x, x, v, v, v, v};
else {
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
}

Node query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u];
else {
int mid = tr[u].l + tr[u].r >> 1;
if (r <= mid) return query(u << 1, l, r);
else if (l > mid) return query(u << 1 | 1, l, r);
else {
auto lnode = query(u << 1, l, r);
auto rnode = query(u << 1 | 1, l, r);
Node res;
pushup(res, lnode, rnode);
return res;
}
}
}

int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build(1, 1, n);

int x, y, k;
while (m--) {
scanf("%d%d%d", &k, &x, &y);
if (k == 1) {
if (x > y) swap(x, y);
auto t = query(1, x, y);
printf("%d\n", t.tmax);
} else {
modify(1, x, y);
}
}
return 0;
}


线段树

1. build，modify，query，pushup，pushdown五种操作
2. 建树的时候，区间大小是确定的，只需要记住区间长度即可

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 200010;
int n, m, p;
struct node {
int l, r;
int v;
} tr[N << 2];

void build(int u, int l, int r) {
tr[u] = {l, r};
if (l == r) return ;
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}

int query(int u, int l, int r) {  // 从u结点开始向下找，在区间l，r中的最大值
if (tr[u].l >= l && tr[u].r <= r) return tr[u].v;
int mid = tr[u].l + tr[u].r >> 1;
int v = 0;
if (l <= mid) v = query(u << 1, l, r);
if (r > mid) v = max(v, query(u << 1 | 1, l, r));
return v;
}

void pushup(int u) {
tr[u].v = max(tr[u << 1].v, tr[u << 1 | 1].v);
}

void modify(int u, int x, int v) {  //从u节点开始向下找序号为x的叶子结点
if (tr[u].l == x && tr[u].r == x) tr[u].v = v;
else {
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
}

int main() {
cin >> m >> p;
build(1, 1, m);

char s[2];
int l, t, last;
while (m--) {
scanf("%s", s);
if (*s == 'Q') {
scanf("%d", &l);
last = query(1, n - l + 1, n);
printf("%d\n", last);
} else {
scanf("%d", &t);
modify(1, n + 1, (t + last) % p);
n++;
}
}

return 0;
}


线段树

1. build，modify，query，pushup，pushdown五种操作
2. 建树的时候，区间大小是确定的，只需要记住区间长度即可

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 200010;
int n, m, p;
struct node {
int l, r;
int v;
} tr[N << 2];

void build(int u, int l, int r) {
tr[u] = {l, r};
if (l == r) return ;
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}

int query(int u, int l, int r) {  // 从u结点开始向下找，在区间l，r中的最大值
if (tr[u].l >= l && tr[u].r <= r) return tr[u].v;
int mid = tr[u].l + tr[u].r >> 1;
int v = 0;
if (l <= mid) v = query(u << 1, l, r);
if (r > mid) v = max(v, query(u << 1 | 1, l, r));
return v;
}

void pushup(int u) {
tr[u].v = max(tr[u << 1].v, tr[u << 1 | 1].v);
}

void modify(int u, int x, int v) {  //从u节点开始向下找序号为x的叶子结点
if (tr[u].l == x && tr[u].r == x) tr[u].v = v;
else {
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
}

int main() {
cin >> m >> p;
build(1, 1, m);

char s[2];
int l, t, last;
while (m--) {
scanf("%s", s);
if (*s == 'Q') {
scanf("%d", &l);
last = query(1, n - l + 1, n);
printf("%d\n", last);
} else {
scanf("%d", &t);
modify(1, n + 1, (t + last) % p);
n++;
}
}

return 0;
}


树状数组

1. 两种操作：单点修改，区间求和，维护一个前缀和序列
2. 想到了如何求牛的身高，但是没能看透两种操作的本质

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
using namespace std;
const int N = 100010;
int n;
int tr[N], a[N], b[N];

int lowbit(int x) {
return x & -x;
}

void add(int x, int c) {
for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

int sum(int x) {
int res = 0;
for (int i = x; i; i -= lowbit(i)) res += tr[i];
return res;
}

int main() {
cin >> n;
a[1] = 0;
for (int i = 2; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= n; i++) {
}
for (int i = n; i; i--) {
int l = 1, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (sum(mid) >= a[i] + 1) r = mid;
else l = mid + 1;
}
b[i] = l;
}
for (int i = 1; i <= n; i++) printf("%d\n", b[i]);

return 0;
}


树状数组

1. 两种操作：单点修改，区间求和，维护一个前缀和序列
2. 想到了如何求牛的身高，但是没能看透两种操作的本质

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
using namespace std;
const int N = 100010;
int n;
int tr[N], a[N], b[N];

int lowbit(int x) {
return x & -x;
}

void add(int x, int c) {
for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}

int sum(int x) {
int res = 0;
for (int i = x; i; i -= lowbit(i)) res += tr[i];
return res;
}

int main() {
cin >> n;
a[1] = 0;
for (int i = 2; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= n; i++) {
}
for (int i = n; i; i--) {
int l = 1, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (sum(mid) >= a[i] + 1) r = mid;
else l = mid + 1;
}
b[i] = l;