码风优化
由于我代码风格极差（使用单字母或缩写或拼音命名，不空格，缩进只用空格不用tab，不写注释），因此导致了打代码时错误不断，难查找问题，导致比赛时出现大量无用时间，因此特此痛下决心改变码风。
参考：C++ 代码规范
测试用题：线段树
原代码
点击查看代码
#include<bits/stdc++.h>
#define for1(i,a,b) for(ll i=a;i<=b;i++)
#define ll long long
#define ls q*2
#define rs q*2+1
using namespace std;
struct node{
ll l;
ll r;
ll lan;
ll w;
}s[500005*4];
ll n,m,a[500005];
void build(ll q,ll l,ll r)
{
s[q].l=l;
s[q].r=r;
s[q].lan=0;
if(l==r)
{
s[q].w=a[l];
return ;
}
ll mid=(l+r)>>1;
build(ls,l,mid);
build(rs,mid+1,r);
s[q].w=s[ls].w+s[rs].w;
}
void chuan(ll q)
{
if(s[q].lan!=0)
{
s[ls].w+=s[q].lan*(s[ls].r-s[ls].l+1);
s[rs].w+=s[q].lan*(s[rs].r-s[rs].l+1);
s[ls].lan+=s[q].lan;
s[rs].lan+=s[q].lan;
s[q].lan=0;
}
return ;
}
void xg(ll q,ll l,ll r,ll k)
{
if(s[q].l>=l&&s[q].r<=r)
{
s[q].w+=k*(s[q].r-s[q].l+1);
s[q].lan+=k;
return ;
}
chuan(q);
ll mid=(s[q].l+s[q].r)>>1;
if(l<=mid) xg(ls,l,r,k);
if(r>mid) xg(rs,l,r,k);
s[q].w=s[ls].w+s[rs].w;
return ;
}
ll cx(ll q,ll l,ll r)
{
// cout<<q<<' '<<s[q].l<<' '<<s[q].r<<endl;
if(s[q].l>=l&&s[q].r<=r)
{
return s[q].w;
}
chuan(q);
ll mid=(s[q].l+s[q].r)>>1;
ll ans=0;
if(l<=mid) ans+=cx(ls,l,r);
if(r>mid) ans+=cx(rs,l,r);
return ans;
}
int main()
{
cin>>n>>m;
for1(i,1,n) scanf("%d",a+i);
ll x,y,k;
build(1,1,n);
for1(i,1,m)
{
ll ji;
cin>>ji;
if(ji==1)
{
cin>>x>>y>>k;
xg(1,x,y,k);
}
else{
cin>>x>>y;
cout<<cx(1,x,y)<<endl;
}
}
return 0;
}
调整后
点击查看代码
#include<bits/stdc++.h>
#define for1(i,a,b) for (ll i = a;i <= b;i ++)
#define ll long long
#define leftSon now*2
#define rightSon now*2+1
const ll N = 1e5;
ll a[N],n,m;
struct SegmentTree
{
ll left[N * 4];
ll right[N * 4];
ll weight[N * 4];
ll lazyTag[N * 4];
SegmentTree()
{
for1(i,1,n*4)
lazyTag[i] = 0;
}
void Build (const ll &now, const ll &l , const ll &r)//建树
{
left[now] = l;
right[now] = r;
if (l == r)
{
weight[now] = a[l];
return ;
}
ll mid = (l + r) / 2;
Build (leftSon, l, mid);
Build (rightSon, mid+1, r);
weight[now] = weight[leftSon] + weight[rightSon];
return ;
}
void Lower (const ll &now)//下放懒标记
{
if(lazyTag[now] == 0) return ;
weight[leftSon] += (right[leftSon] - left[leftSon] + 1) * lazyTag[now];
weight[rightSon] += (right[rightSon] - left[rightSon] + 1) * lazyTag[now];
lazyTag[leftSon] += lazyTag[now];
lazyTag[rightSon] += lazyTag[now];
lazyTag[now]=0;
return ;
}
void Modify (const ll &now, const ll &l, const ll &r, const ll &k)
{
if (left[now] >= l && right[now] <= r)
{
weight[now] += (right[now] - left[now] + 1) * k;
lazyTag[now] += k;
return ;
}
Lower (now);
ll mid = (left[now] + right[now]) / 2;
if (l <= mid) Modify (leftSon, l, r, k);
if (r > mid) Modify (rightSon, l, r, k);
weight[now] = weight[leftSon] + weight[rightSon];
return ;
}
ll Query (const ll &now, const ll &l, const ll &r)//回答询问
{
if (left[now] >= l && right[now] <= r)
return weight[now];
Lower (now);
ll mid = (left[now] + right[now]) / 2;
ll ans=0;
if (l <= mid) ans += Query (leftSon, l, r);
if (r > mid) ans += Query (rightSon, l, r);
return ans;
}
}Tree;
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cin >> n >> m;
for1(i,1,n)
std::cin >> a[i];
Tree.Build (1, 1, n);
ll flag, x, y, k;
for1(i,1,m)
{
std::cin >> flag;
if (flag == 1)
{
std::cin >> x >> y >> k;
Tree.Modify (1, x, y, k);
}
else
{
std::cin >> x >> y;
std::cout<< Tree.Query (1, x, y) <<'\n';
}
}
return 0;
}
测试用题：树状数组
原：
点击查看代码
#include<bits/stdc++.h>
#define for1(i,a,b) for (ll i = a;i <= b;i ++)
#define ll long long
#define leftSon now*2
#define rightSon now*2+1
const ll N = 5e5;
ll a[N],n,m;
struct BinaryIndexedTree
{
int array[N*2];
BinaryIndexedTree()
{
for1(i,1,n*2)
array[i]=0;
}
int lowbit (const int &x)
{
return x&-x;
}
void Modify (const int &l, const int &k)
{
int x=l;
while (x <= n)
{
array[x] += k;
x += lowbit(x);
}
return;
}
int Query (const int &l, const int &r)
{
int x=l-1;
int y=r;
int ans=0;
while(y)
{
ans += array[y];
y -= lowbit(y);
}
while(x)
{
ans -= array[x];
x -= lowbit(x);
}
return ans;
}
void Build (const int &l, const int &r)
{
for1(i,l,r)
Modify (i,a[i]);
}
}Tree;
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cin >> n >> m;
for1(i,1,n)
std::cin >> a[i];
Tree.Build (1, n);
ll flag, x, y, k;
for1(i,1,m)
{
std::cin >> flag;
if (flag == 1)
{
std::cin >> x >> k;
Tree.Modify (x, k);
}
else
{
std::cin >> x >> y;
std::cout<< Tree.Query (x, y) <<'\n';
}
}
return 0;
}
调整后
点击查看代码
#include<bits/stdc++.h>
#define for1(i,a,b) for (ll i = a;i <= b;i ++)
#define ll long long
#define leftSon now*2
#define rightSon now*2+1
const ll N = 5e5+10;
ll a[N],n,m;
struct BinaryIndexedTree
{
int array[N];
BinaryIndexedTree()
{
for1(i,1,n)
array[i]=0;
}
int lowbit (const int &x)
{
return x&-x;
}
void Modify (const int &l, const int &k)
{
int x=l;
while (x <= n)
{
array[x] += k;
x += lowbit(x);
}
return;
}
int Query (const int &l, const int &r)
{
int x=l-1;
int y=r;
int ans=0;
while(y)
{
ans += array[y];
y -= lowbit(y);
}
while(x)
{
ans -= array[x];
x -= lowbit(x);
}
return ans;
}
void Build (const int &l, const int &r)
{
for1(i,l,r)
Modify (i,a[i]);
}
}Tree;
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cin >> n >> m;
for1(i,1,n)
std::cin >> a[i];
Tree.Build (1, n);
ll flag, x, y, k;
for1(i,1,m)
{
std::cin >> flag;
if (flag == 1)
{
std::cin >> x >> k;
Tree.Modify (x, k);
}
else
{
std::cin >> x >> y;
std::cout<< Tree.Query (x, y) <<'\n';
}
}
return 0;
}
在根据新码风的方式打了之后能够明显感觉到的是：
- 很不顺手，经常忘记加空格，std之类的
- 打的时候用时较长
- 可读行明显增强，看起来非常舒服
- 常数明显变小，大概少了100~200ms左右
等用了一段时间之后看看效果如何
于2023/2/4更新
今天线段树合并写了250行，写的人都麻了，一进题解一看其他人同样的做法才80行。。。
感觉以后可以适当的不封装。。。
点击查看代码
#include<bits/stdc++.h>
#define for1(i,a,b) for (int i = a;i <= b;i ++)
#define ll long long
const int N = 1e6+10;
int n, m, Ans[N], R;
struct Side//边
{
int nex;
int to;
Side()
{
int nex = 0;
int to = 0;
}
};
struct Segment//线段
{
int l;
int r;
int t;//属于哪个点的线段树
int d;//权值
Segment()
{
l=r=t=d=0;
}
};
struct Point//点
{
int hd;
int fa;//父亲
int dep;//深度
int rt;//该点的线段树的编号
Point()
{
hd = 0;
fa = 0;
dep = 0;
}
};
struct SegmentTree{
Segment a[N*10];
int cnt2;
SegmentTree()
{
cnt2 = 0;
}
void PushUp (const int x)
{
if(a[a[x].l].d >= a[a[x].r].d)
{
a[x].d = a[a[x].l].d;
a[x].t = a[a[x].l].t;
}
else
{
a[x].d = a[a[x].r].d;
a[x].t = a[a[x].r].t;
}
return ;
}
int Change (const int &ji, const int &x, const int &y, const int &pos, const int &val)
//当前节点，左端点，右端点，修改的位置，修改的权值
{
int now = ji;
if (ji == 0)
now = ++ cnt2;//动态开点
if (x == y)
{
a[now].d += val;
a[now].t = x;
return now;
}
int mid = (x + y) / 2;
if(pos <= mid) a[now].l = Change(a[now].l, x, mid, pos, val);
else a[now].r = Change(a[now].r, mid + 1, y, pos, val);
PushUp(now);
return now;
}
int Merge (int now1, int now2, int x, int y)//线段树合并
{
if (now1 == 0) return now2;
if (now2 == 0) return now1;
if (x == y)
{
a[now1].d += a[now2].d;
a[now1].t = x;
return now1;
}
int mid = (x + y) / 2;
a[now1].l= Merge (a[now1].l, a[now2].l, x, mid);
a[now1].r= Merge (a[now1].r, a[now2].r, mid + 1, y);
PushUp (now1);
return now1;
}
}Tree;
struct ChainTable//链表
{
Point p[N];
Side a[N * 2];
int cnt1;
int st[N][30];
ChainTable()
{
cnt1 = 0;
}
void add (const int &x, const int &y)
{
a[++ cnt1].to = y;
a[cnt1].nex = p[x].hd;
p[x].hd = cnt1;
}
void dfs (const int &x,const int &fa)
{
p[x].dep = p[fa].dep + 1;
p[x].fa = fa;
st[x][0] = fa;
for (int i=1;(1<<i) <= p[x].dep;i++)
st[x][i] = st[st[x][i - 1]][i - 1];
for(int i=p[x].hd;i;i=a[i].nex)
{
int v=a[i].to;
if(v!=fa)
dfs(v,x);
}
}
int Lca (int x, int y)
{
if (p[x].dep > p[y].dep)
std::swap(x,y);
for (int i = 25;i >= 0;i--)
{
if (p[y].dep - (1<<i) >= p[x].dep)
y = st[y][i];
}
if(x==y)
return x;
for(int i = 25;i >= 0;i --)
{
if (st[x][i] != st[y][i])
{
x = st[x][i];
y = st[y][i];
}
}
return st[x][0];
}
void QianZhuiHe (const int &x)
{
for(int i = p[x].hd;i;i = a[i].nex)
{
int v = a[i].to;
if(p[v].dep > p[x].dep)
{
QianZhuiHe(v);
p[x].rt = Tree.Merge(p[x].rt, p[v].rt, 1, R);
}
}
if(Tree.a[p[x].rt].d)
Ans[x] = Tree.a[p[x].rt].t;
}
}lianBiao;
struct Query//询问
{
int x;
int y;
int z;
void Read(const int &l ,const int &r, const int &w)
{
x = l;
y = r;
z = w;
return ;
}
}xunWen[N];
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cin >> n >> m;
int x, y, z;
for1(i,1,n-1)
{
std::cin >> x >> y;
lianBiao.add(x,y);
lianBiao.add(y,x);
}
lianBiao.p[1].dep=1;
lianBiao.dfs (1, 0);
for1(i,1,m)
{
std::cin >> x >> y >> z;
xunWen[i].Read(x,y,z);
R = std::max(R,z);
}
for1(i,1,m)
{
int lca = lianBiao.Lca (xunWen[i].x, xunWen[i].y);
int jx = xunWen[i].x, jy = xunWen[i].y,jz = xunWen[i].z;
int jfa = lianBiao.p[lca].fa;
lianBiao.p[jx].rt = Tree.Change (lianBiao.p[jx].rt, 1, R, jz, 1);
lianBiao.p[jy].rt = Tree.Change (lianBiao.p[jy].rt, 1, R, jz, 1);
lianBiao.p[lca].rt = Tree.Change (lianBiao.p[lca].rt, 1, R, jz, -1);
if (jfa != 0)
lianBiao.p[jfa].rt = Tree.Change (lianBiao.p[jfa].rt, 1, R, jz, -1);
}
lianBiao.QianZhuiHe(1);
for1(i,1,n)
std::cout<< Ans[i] << '\n';
return 0;
return 0;
}
2023/2/7更新
待续未完
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