多校1007 Naive Operations
思路:好像是第一次这么印象深刻的写线段树,说实话,这个题确实很有意思,值得学习。
看了大神讲解视频,但是自己写的还是超时了。
参考来自 https://blog.csdn.net/yiqzq/article/details/81211652 个人认为可以作为模板
#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <cctype>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define MOD 1e9+7
#define PI acos(-1)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + ; int n, q;
int MIN[maxn << ];//记录区间最小值
int lazy[maxn << ];//延迟标记减法
int ans[maxn << ];//记录答案数量
int b[maxn << ];//记录b数组 //线段树基本操作,pushup和pushdown
void pushup(int rt) {
MIN[rt] = min(MIN[rt << ], MIN[rt << | ]);
ans[rt] = ans[rt << ] + ans[rt << | ];
} void pushdown(int rt) {
if(lazy[rt]) {
lazy[rt << ] += lazy[rt];
lazy[rt << | ] += lazy[rt];
MIN[rt << ] -= lazy[rt];
MIN[rt << | ] -= lazy[rt];
lazy[rt] = ;
}
}
//建树
void build(int l, int r, int rt) {
lazy[rt] = ;
ans[rt] = ;
if(l == r) {
scanf("%d", &b[rt]);
MIN[rt] = b[rt];
return ;
}
int m = (l + r) >> ;
build(lson);
build(rson);
pushup(rt);
}
//区间更新,需要注意当NIN减为0的时候,要ans++并且重新将b的值赋给MIN
void updata(int L, int R, int l, int r, int rt) {
if(L <= l && R >= r) {
MIN[rt]--;
if(MIN[rt]) {
lazy[rt]++;
return;
} else {
if(l == r) {
ans[rt]++;
MIN[rt] = b[rt];
return;
/*血的教训,这个return不能放大括号外面,因为如果递归到某个节点MIN是0
但是又不是叶子节点,那么还是需要继续递归直到找到叶子节点为止
*/
} }
}
pushdown(rt);
int m = (l + r) >> ;
if(L <= m) updata(L, R, lson);
if(R > m) updata(L, R, rson);
pushup(rt);
}
//查询
int query(int L, int R, int l, int r, int rt) {
if(L <= l && R >= r) {
return ans[rt];
}
pushdown(rt);
int m = (l + r) >> ;
int sum = ;
if(m >= L) sum += query(L, R, lson);
if(m < R) sum += query(L, R, rson);
return sum;
}
int main() {
while(~scanf("%d%d", &n, &q)) {
build(, n, );
for(int i = ; i <= q; i++) {
char op[];
int a, b;
scanf("%s%d%d", op, &a, &b);
if(op[] == 'q') {
printf("%d\n", query(a, b, , n, ));
} else {
updata(a, b, , n, );
}
}
}
return ;
}
复制来的代码
#include<queue>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm> using namespace std; #define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = 1e5 + ; int add[maxn << ];
int tree[maxn << ];
int MAX[maxn << ];
int a[maxn]; void build(int l, int r,int rt)
{ add[rt] = ;
MAX[rt] = (int)-1e9;
tree[rt] = ; if (l == r){
add[rt] = ;
tree[rt] = ;
MAX[rt] = -a[l];
return;
}
int m = (l + r) >> ;
build(lson);
build(rson);
tree[rt] = tree[rt << ] + tree[rt << | ];
MAX[rt] = max(MAX[rt << ], MAX[rt << | ]);
}
void PushDown(int rt)
{
if (add[rt]){
add[rt << ] += add[rt];
add[rt << | ] += add[rt];
MAX[rt << ] += add[rt];
MAX[rt << | ] += add[rt];
add[rt] = ;
}
}
void update1(int L, int R, int l, int r, int rt)
{
if (L <= l&&r <= R){
add[rt]++;
MAX[rt]++;
return;
}
PushDown(rt);
int m = (l + r) >> ;
if (L <= m)update1(L, R, lson);
if (R > m)update1(L, R, rson);
tree[rt] = tree[rt << ] + tree[rt << | ];
MAX[rt] = max(MAX[rt << ], MAX[rt << | ]);
}
void update2(int L, int R, int l, int r, int rt)
{
if (l == r){ if (MAX[rt] >= ){ tree[rt]++;
MAX[rt] = -a[l];
}
return;
}
PushDown(rt);
int m = (l + r) >> ;
if (L <= m&&MAX[rt << ] >= )update2(L, R, lson);
if (R > m&&MAX[rt << | ] >= )update2(L, R, rson);
tree[rt] = tree[rt << ] + tree[rt << | ];
MAX[rt] = max(MAX[rt << ], MAX[rt << | ]);
}
int query(int L, int R, int l, int r, int rt)
{
if (L <= l&&r <= R){
return tree[rt];
}
int m = (l + r) >> , ans = ;
if (L <= m)ans += query(L, R, lson);
if (R > m)ans += query(L, R, rson);
return ans;
}
void Print(int l, int r, int rt)
{
if (l == r){
cout << rt << " = " << MAX[rt] << endl;
return;
}
cout << rt << " = " << MAX[rt] << endl;
int m = (l + r) >> ;
if (l <= m)Print(lson);
if (r > m)Print(rson);
}
int main()
{
std::ios::sync_with_stdio(false);
int n, q;
while (cin >> n >> q){
for (int i = ; i <= n; i++)
cin >> a[i]; build(, n, ); string flag; int l, r;
for (int i = ; i < q; i++){
cin >> flag >> l >> r; if (flag == "add"){
update1(l, r, , n, );
update2(l, r, , n, );
//Print(1, n, 1);
}
else cout << query(l, r, , n, ) << endl;
}
} return ;
}
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