B题，数据范围1e5，范围较小。
区间修改，区间查询，很明显的线段树+懒标记板子,套一块模板即可
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, w[N];
struct node {
int l, r;
int sum, add;
} tr[4 * N];
void pushup(int u) {
tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}
void pushdown(int u) {
auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
if (root.add) {
left.add = max(root.add, left.add);
right.add = max(root.add, right.add);
root.add = 0;
}
}
void build(int u, int l, int r) {
if (l == r) {
tr[u] = {
l, r, w[r], 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 = max(d, tr[u].add);
} else {
pushdown(u);
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);
}
}
int query(int u, int l, int r) {
if (tr[u].l == tr[u].r) {
return max(tr[u].sum,tr[u].add);
}
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
int sum = 0;
if (l <= mid)
sum = query(u << 1, l, r);
if (r > mid)
sum += query(u << 1 | 1, l, r);
return sum;
}
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
char ch;
cin >> ch;
w[i] = ch;
}
build(1, 1, n);
while (m--) {
int l, r;
char ch;
cin >> l >> r >> ch;
int k = ch;
modify(1, l, r, k);
}
cout << query(1, 1, n) << endl;
}
也可以用类差分+优先队列的方法来做,将覆盖区间的左右端点打上标记，当我们从左到右扫描时遇到开始标记，就将这个标记纳入优先队列，同时设个数组进行标记，当遇到结束标记时，就将这个魔法在标记数组中减1，同时，如果优先队列的顶部的标记消失了，就需要弹出。
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef pair<int, int>PII;
const int N = 1e5 + 10;
int n, m;
int st[200];//哪些是捏在手里的
priority_queue <char, vector<char>, less<char> >q; //现在正在用哪些东西在扫
char a[N];
vector<char>gett[N], lose[N];
void solve() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
char ch;
cin >> a[i];
}
while (m--) {
int l, r;
char ch;
cin >> l >> r >> ch;
gett[l].push_back(ch);
lose[r + 1].push_back(ch);
}
int ans = 0;
for (int i = 1; i <= n; i++) {
for (auto j : gett[i]) {
q.push(j);
st[j]++;
}
for (auto j : lose[i]) {
st[j]--;
while (q.size() && st[q.top()] == 0) {
q.pop();
}
}
if (q.size())
a[i] = max(a[i], q.top());
ans += a[i];
}
cout << ans << endl;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
}
C题，数据范围1e7，线段树和差分都会爆空间，考虑用并查集加速:
将所有魔法按优先级sort一下，然后暴力填涂，如果已经被涂过了就利用并查集跳到未填涂的地方，否则就在填涂该格后并查集指向区间的右端
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
int n, m;
const int N = 1e7 + 10;
int f[N];//指向一个区间里的主线
struct node {
char val;
int l, r;
} q[N];
char a[N];
bool cmp(node a, node b) {
if (a.val == b.val) {
return a.l < b.l;
}
return a.val > b.val;
}
int find(int x) {
return x == f[x] ? f[x] : find(f[x]);
}
void solve() {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> a[i];
f[i] = i;
}
for (int i = 1; i <= m; i++) {
cin >> q[i].l >> q[i].r >> q[i].val;
}
sort(q + 1, q + 1 + m, cmp);
for (int i = 1; i <= m; i++) {
int l = q[i].l, r = q[i].r;
char ch = q[i].val;
if(find(l)>r)continue;
for (int i = l; i <= r; i++) {
i = find(i);
if (i > r)
break;
f[i] = max(f[i], r);
a[i] = max(a[i], ch);
}
}
int ans = 0;
for (int i = 1; i <= n; i++) {
ans += a[i];
}
cout << ans << endl;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
}