wenbao与RMP（线段树）

Posted 2020-08-08 wenbao

rmq

就是在变化的前提下在一定范围内查询最值

 

@单点更新

http://hihocoder.com/problemset/problem/1077

在大神面前这就是水题？？？？

自愧不如

 

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 const int maxn = 1e6+10;
 5 #define INF  0x3f3f3f3f; 
 6 struct Node{
 7     int lx, rx, mi;
 8 } Tree[maxn<<2];
 9 
10 void build(int lx, int rx, int node){
11     Tree[node].lx = lx;
12     Tree[node].rx = rx;
13     if(lx == rx){
14         scanf("%d", &Tree[node].mi);
15         return ;
16     }
17     int mid = (lx + rx) >> 1;
18     build(lx, mid, node<<1);
19     build(mid+1, rx, node<<1|1);
20     Tree[node].mi = min(Tree[node<<1].mi, Tree[node<<1|1].mi);
21 }
22 
23 void update(int point, int v, int node){
24     if(Tree[node].lx == Tree[node].rx){
25         Tree[node].mi = v;
26         return ;
27     }
28     if(point >= Tree[node<<1|1].lx) update(point, v, node<<1|1);
29     if(point <= Tree[node<<1].rx) update(point, v, node<<1);
30     Tree[node].mi = min(Tree[node<<1].mi, Tree[node<<1|1].mi);
31 }
32 
33 int query(int lx, int rx, int node){
34     if(Tree[node].lx >= lx && Tree[node].rx <= rx){
35         return Tree[node].mi;
36     }
37     int a = INF;
38     int b = INF;
39     if(lx <= Tree[node<<1].rx) a = query(lx, rx, node<<1);
40     if(rx >= Tree[node<<1|1].lx) b = query(lx, rx, node<<1|1);
41     return min(a, b);
42 }
43 
44 int main(){
45     int n, m, a, b, c;
46     scanf("%d", &n);
47     build(1, n, 1);
48     cin>>m;
49     for(int i = 0; i < m; i++){
50         scanf("%d %d %d", &a, &b, &c);
51         if(a){
52             update(b, c, 1);
53         }
54         else{
55             printf("%d\n",query(b, c, 1));
56         }
57     }
58     return 0;
59 }

 

@区间更新

http://hihocoder.com/problemset/problem/1078

学习到了。。。。。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 const int maxn = 1e5 + 10;
 5 int sum[maxn << 2], lazy[maxn << 2]; 
 6 #define lson lx, mid, node << 1
 7 #define rson mid + 1, rx, node << 1 | 1
 8 
 9 void pushup(int node){
10     sum[node] = sum[node << 1] + sum[node << 1 | 1]; 
11 }
12 
13 void pushdown(int node, int rangl, int rangr){
14     if(lazy[node]){
15         sum[node << 1] = rangl * lazy[node];    //。。。。。。。。。。。。。。。
16         sum[node << 1 | 1] = rangr * lazy[node];    //此处写错了，調了半天
17         lazy[node << 1] = lazy[node];
18         lazy[node << 1 | 1] = lazy[node];
19         lazy[node] = 0;
20     }
21 }
22 
23 void build(int lx, int rx, int node){
24     lazy[node] = 0;
25     if(lx == rx){
26         scanf("%d", &sum[node]);
27         return ;
28     }
29     int mid = (lx + rx) >>1;
30     build(lson);
31     build(rson);
32     pushup(node);
33 }
34 
35 void update(int xl, int xr, int val, int lx, int rx, int node){
36     if(xl <= lx && rx <= xr){
37         sum[node] = (rx - lx + 1) *val;
38         lazy[node] = val;
39         return ;
40     }
41     int mid = (lx + rx) >> 1;
42     pushdown(node, mid - lx + 1, rx - mid);
43     if(xl <= mid) update(xl, xr, val, lson);
44     if(mid < xr) update(xl, xr, val, rson);
45     pushup(node);
46 }
47 
48 int query(int xl, int xr, int lx, int rx, int node){
49     if(xl <= lx && rx <= xr){
50         return sum[node];
51     }
52     int ans = 0;
53     int mid = (lx + rx) >>1;
54     pushdown(node, mid - lx + 1, rx - mid);
55     if(xl <= mid)  ans += query(xl, xr, lson);
56     if(mid < xr) ans += query(xl, xr, rson);
57     return ans;
58 }
59 
60 int main(){
61     int n, m, a, b, c, d;
62     scanf("%d", &n);
63     build(1, n, 1);
64     scanf("%d", &m);
65     while(m--){
66         scanf("%d", &a);
67         if(a){
68             scanf("%d %d %d", &b, &c, &d);
69             update(b, c, d, 1, n, 1);
70         }
71         else{
72             scanf("%d %d", &b, &c);
73             printf("%d\n", query(b, c, 1, n, 1));
74         }
75     }
76     return 0;
77 }