hdu4348 - To the moon 可持久化线段树 区间修改 离线处理

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

 

posted @ 2016-03-02 08:28  Showson  阅读(286)  评论(0)  编辑  收藏  举报