标记下模板，需要的时候抄写
1.线段树 单点更新，区间查询
https://hihocoder.com/contest/offers27/problem/3
1 #include<bits/stdc++.h> 2 #define pb push_back 3 typedef long long ll; 4 using namespace std; 5 typedef pair<int, int> pii; 6 const int maxn = 2e5 + 10; 7 int f[maxn * 4]; 8 int a[maxn]; 9 vector<int> e[100010]; 10 int n; 11 int ti[maxn]; 12 int cnt = 0; 13 int rt[100010][2]; 14 void dfs(int u, int p) { 15 ti[++cnt] = u; 16 rt[u][0] = cnt; 17 for (int x : e[u]) { 18 if(x == p) continue; 19 dfs(x, u); 20 } 21 ti[++cnt] = u; 22 rt[u][1] = cnt; 23 } 24 void bt(int x, int y, int o) { 25 if(x == y) { 26 f[o] = a[ti[x]]; 27 return; 28 } 29 int mid = (x + y) >> 1; 30 bt(x, mid, o * 2); 31 bt(mid + 1, y, o * 2 + 1); 32 f[o] = min(f[o * 2], f[o * 2 + 1]); 33 } 34 int tx1, ty1; 35 int ask(int x, int y, int o) { 36 if(x >= tx1 && y <= ty1) return f[o]; 37 int mid = (x + y) >> 1; 38 int res = INT_MAX; 39 if(tx1 <= mid) 40 res = min(res, ask(x, mid, o * 2)); 41 if(ty1 >= mid + 1) 42 res = min(res, ask(mid + 1, y, o * 2 + 1)); 43 return res; 44 } 45 void up(int x, int y, int o) { 46 if(y < tx1 || x > tx1) return; 47 if(x == y && x == tx1) { 48 f[o] = ty1; 49 //cout << o << " asd " << f[o] << " " << y1 << endl; 50 return; 51 } 52 int mid = (x + y) >> 1; 53 if(tx1 <= mid) up(x, mid, o * 2); 54 else up(mid + 1, y, o * 2 + 1); 55 f[o] = min(f[o * 2], f[o * 2 + 1]); 56 //cout << o << " asd " << f[o] << endl; 57 } 58 void solve() { 59 scanf("%d", &n); 60 for (int i = 1; i <= n; i++) scanf("%d", &a[i]); 61 int x, y; 62 for (int i = 2; i <= n; i++) { 63 scanf("%d", &x); 64 e[x].pb(i); 65 } 66 dfs(1, 0); 67 bt(1, cnt, 1); 68 int m; 69 scanf("%d", &m); 70 while(m--) { 71 scanf("%d%d", &x, &y); 72 //cout << x << " " << y << endl; 73 if(x == 1) { 74 scanf("%d", &x); 75 swap(x, y); 76 tx1 = rt[x][0]; 77 ty1 = y; 78 up(1, cnt, 1); 79 tx1 = rt[x][1]; 80 up(1, cnt, 1); 81 82 } else { 83 tx1 = rt[y][0]; ty1 = rt[y][1]; 84 int t = ask(1, cnt, 1); 85 printf("%d ", t); 86 } 87 } 88 89 } 90 91 int main() { 92 //freopen("test.in", "r", stdin); 93 //freopen("test.out", "w", stdout); 94 solve(); 95 return 0; 96 }
2. 区间更新，区间查询， lazy标记
https://hihocoder.com/contest/offers28/problem/3
1 #include<bits/stdc++.h> 2 #define pb push_back 3 typedef long long ll; 4 using namespace std; 5 typedef pair<int, int> pii; 6 const int maxn = 1e5 + 10; 7 8 int n, m, q; 9 pii a[maxn], b[maxn]; 10 11 //tree 12 int x, y, v; 13 int sum[1863000],tag[1863000]; 14 #define pd(o,l,r) tag[o<<1] += tag[o], tag[o<<1|1]+=tag[o],sum[o<<1]+=tag[o],sum[o<<1|1]+=tag[o],tag[o]=0 15 #define mid (l+r>>1) 16 void update(int o,int l,int r){ 17 //cout << o << " " << l << " " << r << endl; 18 x<=l&&r<=y? 19 tag[o]+=v,sum[o]=sum[o] + v: 20 (tag[o]?pd(o,l,r),1:1,x<=mid?update(o<<1,l,mid),1:1,mid<y?update(o<<1|1,mid+1,r),1:1,sum[o]= max(sum[o<<1], sum[o<<1|1])); 21 }int query(int o,int l,int r){ 22 int t = 0; 23 x<=l&&r<=y?t = sum[o]:(tag[o]?pd(o,l,r),1:1, 24 x<=mid?t = max(t, query(o<<1,l,mid)),1:1,mid<y?t = max(t, query(o<<1|1,mid+1,r)),1:1); 25 return t; 26 } 27 void solve() { 28 scanf("%d%d", &n, &m); 29 int tx, ty; 30 set<int> se; 31 map<int, int> ma; 32 for (int i = 0; i < n; i++) { 33 scanf("%d%d", &tx, &ty); 34 a[i] = {tx, ty}; se.insert(tx); se.insert(ty); 35 } 36 scanf("%d", &q); 37 for (int i = 0; i < q; i++) { 38 scanf("%d%d", &tx, &ty); 39 b[i] = {tx, ty}; se.insert(tx); se.insert(ty); 40 } 41 int cnt = 0; 42 for (int t : se) { 43 ma[t] = ++cnt; 44 } 45 for (int i = 0; i < n; i++) { 46 x = ma[a[i].first], y = ma[a[i].second]-1; 47 //cout << x << " " << y << endl; 48 v = 1; 49 50 update(1, 1, cnt);//cout << "asd" << endl; 51 } 52 for (int i = 0; i < q; i++) { 53 x = ma[b[i].first], y = ma[b[i].second]-1; 54 //cout << x << " " << y << endl; 55 int t = query(1, 1, cnt) + 1; 56 if(t > m) puts("NO"); 57 else puts("YES"); 58 } 59 } 60 61 int main() { 62 //freopen("test.in", "r", stdin); 63 //freopen("test.out", "w", stdout); 64 solve(); 65 return 0; 66 }
https://hihocoder.com/contest/hiho20/problem/1
1 #include<cstdio> 2 #define mid (l+r>>1) 3 #define pd(o,l,r) tag[o<<1]=tag[o<<1|1]=tag[o],sum[o<<1]=tag[o]*(mid-l+1),sum[o<<1|1]=tag[o]*(r-mid),tag[o]=0 4 int ans,sum[263000],tag[263000],v,n,q,op,x,y,i,aa;char ch;int F(){ 5 while(ch=getchar(),ch<'0'||ch>'9');aa=ch-48; 6 while(ch=getchar(),ch>='0'&&ch<='9')aa=aa*10+ch-48;return aa; 7 }void bt(int o,int l,int r){ 8 l==r?sum[o]=F():(bt(o<<1,l,mid),bt(o<<1|1,mid+1,r),sum[o]=sum[o<<1]+sum[o<<1|1]); 9 }void update(int o,int l,int r){ 10 x<=l&&r<=y?tag[o]=v,sum[o]=v*(r-l+1):(tag[o]?pd(o,l,r),1:1, 11 x<=mid?update(o<<1,l,mid),1:1,mid<y?update(o<<1|1,mid+1,r),1:1,sum[o]=sum[o<<1]+sum[o<<1|1]); 12 }void query(int o,int l,int r){ 13 x<=l&&r<=y?ans+=sum[o]:(tag[o]?pd(o,l,r),1:1, 14 x<=mid?query(o<<1,l,mid),1:1,mid<y?query(o<<1|1,mid+1,r),1:1); 15 }int main(){ 16 for(bt(1,1,n=F()),q=F();q--;op=F(),x=F(),y=F(), 17 op?v=F(),update(1,1,n),1:(ans=0,query(1,1,n),printf("%d ",ans))); 18 }
上面这个代码是十佳代码代码里面的，我一般都是学的这个模板。