TYVJ 1039 忠诚2 by C++
1: #include<iostream>
2: #include<string.h>
3: #include<math.h>
4: using namespace std;
5: int m,n;
6: const int maxn=100000;
7: int w[maxn],f[maxn*50];
8: void get_prepare()
9: {
10: memset(w,0,sizeof(w));
11: memset(f,0,sizeof(f));
12: cin >> m >> n;
13: for (int i=1;i<=m;i++) cin >> w[i];
14: }
15: void build_tree(int a,int b,int p)
16: {
17: if (a==b){
18: f[p]=w[a];
19: return;
20: }
21: int k=(a+b)/2;
22: build_tree(a,k,p*2);
23: build_tree(k+1,b,p*2+1);
24: f[p]=min(f[p*2],f[p*2+1]);
25: }
26: int ask_tree(int a,int b,int l,int r,int p)
27: {
28: if (a==l && b==r) return f[p];
29: int k=(l+r)/2;
30: if (b<=k) return ask_tree(a,b,l,k,p*2);
31: if (a>k) return ask_tree(a,b,k+1,r,p*2+1);
32: return min(ask_tree(a,k,l,k,p*2),ask_tree(k+1,b,k+1,r,p*2+1));
33: }
34: void update(int a,int l,int r,int p)
35: {
36: if (l==r){
37: f[p]=w[l];
38: return;
39: }
40: int k=(l+r)/2;
41: if (a<=k) update(a,l,k,p*2);
42: else update(a,k+1,r,p*2+1);
43: f[p]=min(f[p*2],f[p*2+1]);
44: }
45: void change(int x,int y){
46: w[x]=y;
47: update(x,1,m,1);
48: }
49: int main()
50: {
51: get_prepare();
52: build_tree(1,m,1);
53: int x,y,p;
54: for (int i=0;i<n;i++){
55: cin >> p >> x >> y;
56: if (p==1) cout << ask_tree(x,y,1,m,1) << ' ';
57: else change(x,y);
58: }
59: cout << endl;
60: return 0;
61: }
1. 我想说 在main()里的 那个if-else是被迫改的，原本是p==1?cout << ask_tree(x,y,1,m,1) << ' ':change(x,y);可是不知道为什么在自己机器可以运行，交到tyvj会CE，为通过率泪奔……
2. 对于这个线段树的思想… 不阐述了，好歹没有忘掉哈
3. 我把树的数组开到5*10^6才把最后后两组数据过掉， 再次泪奔……
