法一：暴力！
让干什么就干什么，那么久需要可持久化线段树了。
但是空间好紧。怎么破？
不down标记好了！
每个点维护sum和add两个信息，sum是这段真实的和，add是这段整体加了多少，如果这段区间被完全包含，返回sum，否则加上add * 询问落在这段区间的长度再递归回答。
怎么还是MLE？
麻辣鸡指针好像8字节，所以改成数组的就过了。。。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
using namespace std;
template<typename Q> Q &read(Q &x) {
static char c, f;
for(f = 0; c = getchar(), !isdigit(c); ) if(c == ‘-‘) f = 1;
for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - ‘0‘;
if(f) x = -x; return x;
}
template<typename Q> Q read() {
static Q x; read(x); return x;
}
typedef long long LL;
const int N = 100000 + 10;
struct Node *pis;
struct Node {
LL sum, add;
Node *ch[2];
Node *modify(int l, int r, int L, int R, LL d) {
Node *o = new Node(*this);
if(L <= l && r <= R) {
o->add += d;
o->sum += (r - l + 1) * d;
return o;
}
int mid = (l + r) >> 1;
if(L <= mid) o->ch[0] = ch[0]->modify(l, mid, L, R, d);
if(mid < R) o->ch[1] = ch[1]->modify(mid + 1, r, L, R, d);
o->sum = o->ch[0]->sum + o->ch[1]->sum + o->add * (r - l + 1);
return o;
}
LL query(int l, int r, int L, int R) {
if(L <= l && r <= R) return sum;
int mid = (l + r) >> 1;
LL res = (min(R, r) - max(L, l) + 1) * add;
if(L <= mid) res += ch[0]->query(l, mid, L, R);
if(mid < R) res += ch[1]->query(mid + 1, r, L, R);
return res;
}
void *operator new(size_t) {
return pis++;
}
}pool[1000000 + 10], *root[N];
void build(Node *&o, int l, int r) {
o = new Node, o->add = 0;
if(l == r) return read(o->sum), void();
int mid = (l + r) >> 1;
build(o->ch[0], l, mid);
build(o->ch[1], mid + 1, r);
o->sum = o->ch[0]->sum + o->ch[1]->sum;
}
int main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int n, m, cur;
char opt[10];
while(scanf("%d%d", &n, &m) == 2) {
cur = 0, pis = pool;
build(root[cur], 1, n);
while(m--) {
if(m == 1) {
int debug = 1;
}
scanf("%s", opt);
if(opt[0] == ‘C‘) {
int l, r; LL d;
read(l), read(r), read(d);
root[cur + 1] = root[cur]->modify(1, n, l, r, d);
cur++;
}else if(opt[0] == ‘Q‘) {
int l, r; read(l), read(r);
printf("%I64d\n", root[cur]->query(1, n, l, r));
}else if(opt[0] == ‘H‘) {
int l, r, t; read(l), read(r), read(t);
printf("%I64d\n", root[t]->query(1, n, l, r));
}else read(cur);
}
puts("");
}
return 0;
}
指针版
1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<algorithm>
5 #include<iostream>
6
7 using namespace std;
8
9 template<typename Q> Q &read(Q &x) {
10 static char c, f;
11 for(f = 0; c = getchar(), !isdigit(c); ) if(c == ‘-‘) f = 1;
12 for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - ‘0‘;
13 if(f) x = -x; return x;
14 }
15 template<typename Q> Q read() {
16 static Q x; read(x); return x;
17 }
18
19 typedef long long LL;
20 const int N = 4000000 + 10;
21
22 int ch[N][2], tot, root[N];
23 LL sum[N], add[N];
24
25 int modify(int s, int l, int r, int L, int R, LL d) {
26 int x = tot++;
27 sum[x] = sum[s];
28 add[x] = add[s];
29 ch[x][0] = ch[s][0];
30 ch[x][1] = ch[s][1];
31
32 if(L <= l && r <= R) {
33 add[x] += d;
34 sum[x] += (r - l + 1) * d;
35 }else {
36 int mid = (l + r) >> 1;
37 if(L <= mid) ch[x][0] = modify(ch[s][0], l, mid, L, R, d);
38 if(mid < R) ch[x][1] = modify(ch[s][1], mid + 1, r, L, R, d);
39 sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + add[x] * (r - l + 1);
40 }
41 return x;
42 }
43
44 LL query(int s, int l, int r, int L, int R) {
45 if(L <= l && r <= R) return sum[s];
46 int mid = (l + r) >> 1;
47 LL res = (min(R, r) - max(L, l) + 1) * add[s];
48 if(L <= mid) res += query(ch[s][0], l, mid, L, R);
49 if(mid < R) res += query(ch[s][1], mid + 1, r, L, R);
50 return res;
51 }
52
53 void build(int &s, int l, int r) {
54 s = tot++, add[s] = 0;
55 if(l == r) return read(sum[s]), void();
56 int mid = (l + r) >> 1;
57 build(ch[s][0], l, mid);
58 build(ch[s][1], mid + 1, r);
59 sum[s] = sum[ch[s][0]] + sum[ch[s][1]];
60 }
61
62 int main() {
63 #ifdef DEBUG
64 freopen("in.txt", "r", stdin);
65 freopen("out.txt", "w", stdout);
66 #endif
67
68 int n, m, cur;
69 char opt[10];
70 while(scanf("%d%d", &n, &m) == 2) {
71 cur = tot = 0;
72 build(root[cur], 1, n);
73 while(m--) {
74 if(m == 1) {
75 int debug = 1;
76 }
77 scanf("%s", opt);
78 if(opt[0] == ‘C‘) {
79 int l, r; LL d;
80 read(l), read(r), read(d);
81 root[cur + 1] = modify(root[cur], 1, n, l, r, d);
82 cur++;
83 }else if(opt[0] == ‘Q‘) {
84 int l, r; read(l), read(r);
85 printf("%I64d\n", query(root[cur], 1, n, l, r));
86 }else if(opt[0] == ‘H‘) {
87 int l, r, t; read(l), read(r), read(t);
88 printf("%I64d\n", query(root[t], 1, n, l, r));
89 }else read(cur);
90 }
91 // puts("");
92 }
93
94 return 0;
95 }
数组版
法二：离线！
主要需要处理H操作。
在第一遍读入数据的时候维护一个pos[]数组，表示当前第i个版本是由pos[i]这个C操作创建的。
然后碰到H就把它挂在pos[t]上就可以，第二遍处理的时候直接回答。
1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<algorithm>
5 #include<iostream>
6
7 using namespace std;
8
9 template<typename Q> Q &read(Q &x) {
10 static char c, f;
11 for(f = 0; c = getchar(), !isdigit(c); ) if(c == ‘-‘) f = 1;
12 for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - ‘0‘;
13 if(f) x = -x; return x;
14 }
15 template<typename Q> Q read() {
16 static Q x; read(x); return x;
17 }
18
19 typedef long long LL;
20 const int N = 100000 + 10;
21
22 int n, m;
23 class SegementTree {
24 private:
25 LL sum[N * 4], tag[N * 4];
26
27 #define mid ((l + r) >> 1)
28 #define ls s << 1, l, mid
29 #define rs s << 1 | 1, mid + 1, r
30
31 void add_tag(int s, int l, int r, LL d) {
32 tag[s] += d;
33 sum[s] += (r - l + 1) * d;
34 }
35
36 void down(int s, int l, int r) {
37 if(tag[s]) {
38 add_tag(ls, tag[s]);
39 add_tag(rs, tag[s]);
40 tag[s] = 0;
41 }
42 }
43
44 int lft, rgt;
45 LL w;
46
47 void modify(int s, int l, int r) {
48 if(lft <= l && r <= rgt) return add_tag(s, l, r, w);
49 down(s, l, r);
50 if(lft <= mid) modify(ls);
51 if(mid < rgt) modify(rs);
52 sum[s] = sum[s << 1] + sum[s << 1 | 1];
53 }
54
55 LL query(int s, int l, int r) {
56 if(lft <= l && r <= rgt) return sum[s];
57 down(s, l, r);
58 if(rgt <= mid) return query(ls);
59 if(mid < lft) return query(rs);
60 return query(ls) + query(rs);
61 }
62
63 public:
64 void build(int s, int l, int r) {
65 tag[s] = 0;
66 if(l == r) return read(sum[s]), void();
67 build(ls), build(rs);
68 sum[s] = sum[s << 1] + sum[s << 1 | 1];
69 }
70 #undef mid
71 #undef ls
72 #undef rs
73
74 void Modify(int l, int r, LL w) {
75 lft = l, rgt = r, this->w = w;
76 modify(1, 1, n);
77 }
78 LL Query(int l, int r) {
79 lft = l, rgt = r;
80 return query(1, 1, n);
81 }
82 }seg;
83
84 struct operation {
85 char tp;
86 int l, r;
87 LL d;
88 }opt[N];
89
90 #include<stack>
91 stack<int> stk;
92
93 #include<vector>
94 vector<int> G[N];
95
96 int pos[N];
97 LL ans[N];
98
99 int main() {
100 #ifdef DEBUG
101 freopen("in.txt", "r", stdin);
102 freopen("out.txt", "w", stdout);
103 #endif
104
105 char s[10];
106 while(scanf("%d%d", &n, &m) == 2) {
107 seg.build(1, 1, n);
108 int cur = 0;
109 for(int i = 0; i < m; i++) {
110 scanf("%s", s);
111 opt[i].tp = s[0];
112 if(s[0] == ‘C‘) {
113 read(opt[i].l), read(opt[i].r), read(opt[i].d);
114 pos[++cur] = i;
115 }else if(s[0] == ‘Q‘) {
116 read(opt[i].l), read(opt[i].r);
117 }else if(s[0] == ‘H‘) {
118 read(opt[i].l), read(opt[i].r), read(opt[i].d);
119 if(!opt[i].d) ans[i] = seg.Query(opt[i].l, opt[i].r);
120 else G[pos[opt[i].d]].push_back(i);
121 }else cur = read(opt[i].d);
122 }
123
124 cur = 0;
125 for(int i = 0; i < m; i++) {
126 if(opt[i].tp == ‘C‘) {
127 seg.Modify(opt[i].l, opt[i].r, opt[i].d);
128 for(unsigned j = 0; j < G[i].size(); j++) {
129 int k = G[i][j];
130 ans[k] = seg.Query(opt[k].l, opt[k].r);
131 }
132 ++cur;
133 stk.push(i);
134 }else if(opt[i].tp == ‘Q‘) {
135 ans[i] = seg.Query(opt[i].l, opt[i].r);
136 }else if(opt[i].tp == ‘B‘) {
137 while(cur > opt[i].d) {
138 int k = stk.top(); stk.pop();
139 seg.Modify(opt[k].l, opt[k].r, -opt[k].d);
140 cur--;
141 }
142 }
143 }
144
145 for(int i = 0; i < m; i++) {
146 if(opt[i].tp == ‘Q‘ || opt[i].tp == ‘H‘) {
147 printf("%I64d\n", ans[i]);
148 }
149 }
150 }
151
152 return 0;
153 }
离线版
时间： 2024-12-17 22:10:56