求问用scanf WA92分
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求问用scanf WA92分
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Cap1taL楼主2022/8/3 19:15

只WA了一个点

代码

// Problem: P3038 [USACO11DEC]Grass Planting G
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3038
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#include <bits/extc++.h>
#define INF 0x7fffffff
#define MAXN 200005
#define MAXM 100003
#define eps 1e-9
#define foru(a,b,c)	for(int a=b;a<=c;a++)
#define ford(a,b,c)	for(int a=b;a>=c;a--)
#define RT return 0;
#define db(x)	cout<<endl<<x<<endl;
#define LL long long
#define LXF int
#define RIN rin()
#define HH printf("\n")
using namespace std;
inline LXF rin(){
	LXF x=0,w=1;
	char ch=0;
	while(ch<'0'||ch>'9'){ 
	if(ch=='-') w=-1;
	ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
	x=x*10+(ch-'0');
	ch=getchar();
	}
	return x*w;
}
inline void out(LXF x){
	if(x<0){
	x=-x;
	putchar('-');
	}
	if(x>9) out(x/10);
	putchar(x%10+'0');
}
int n,m,a[MAXN],id[MAXN],dep[MAXN],top[MAXN],fa[MAXN],siz[MAXN],son[MAXN],cnt,ecnt,head[MAXN];
struct E{
	int v,nxt;
}e[MAXM<<1];
void add_e(int u,int v){
	e[++ecnt]=(E){v,head[u]};
	head[u]=ecnt;
}
void dfs1(int s,int fath){
	fa[s]=fath;
	siz[s]=1;
	dep[s]=dep[fath]+1;
	int maxson=-1;
	for(int i=head[s];i;i=e[i].nxt){
		int v=e[i].v;
		if(v!=fath){
			dfs1(v,s);
			siz[s]+=siz[v];
			if(siz[v]>maxson){
				maxson=siz[v];
				son[s]=v;
			}
		}
	}
}
void dfs2(int s,int topf){
	id[s]=++cnt;
	top[s]=topf;
	if(!son[s])	return ;
	dfs2(son[s],topf);
	for(int i=head[s];i;i=e[i].nxt){
		int v=e[i].v;
		if(v!=fa[s]&&v!=son[s]){
			dfs2(v,v);
		}
	}
}
int LCA(int x,int y){
	while(top[x]!=top[y]){
		if(dep[top[x]]<dep[top[y]])	swap(x,y);
		x=fa[top[x]];
	}
	if(dep[x]<dep[y])	swap(x,y);
	return y;
}
//SegTree
struct Segtree{
	int l,r,sum,lz;
}tr[MAXN<<2];
inline int lc(int x){return x<<1;}
inline int rc(int x){return 1+(x<<1);}
void push_up(int p){
	tr[p].sum=tr[lc(p)].sum+tr[rc(p)].sum;
}
void update(int p,int k){
	tr[p].sum+=(tr[p].r-tr[p].l+1)*k;
	tr[p].lz+=k;
}
void push_down(int p){
	update(lc(p),tr[p].lz);
	update(rc(p),tr[p].lz);
	tr[p].lz=0;
}
void build(int p,int l,int r){
	tr[p].l=l,tr[p].r=r;
	if(l==r)	return ;
	int mid=(l+r)>>1;
	build(lc(p),l,mid);
	build(rc(p),mid+1,r);
	push_up(p);
}
void modify(int p,int nl,int nr,int k){
	int l=tr[p].l,r=tr[p].r;
	if(r<nl||nr<l)	return ;
	if(nl<=l&&r<=nr){
		update(p,k);
		return ;
	}
	push_down(p);
	modify(lc(p),nl,nr,k);
	modify(rc(p),nl,nr,k);
	push_up(p);
}
int qurey(int p,int nl,int nr){
	int l=tr[p].l,r=tr[p].r;
	if(r<nl||nr<l)	return 0;
	if(nl<=l&&r<=nr)	return tr[p].sum;
	push_down(p);
	return qurey(lc(p),nl,nr)+qurey(rc(p),nl,nr);
}
void updRange(int x,int y,int k){
	while(top[x]!=top[y]){
		if(dep[top[x]]<dep[top[y]])	swap(x,y);
		// printf("modify(1,%d,%d,%d)\n",id[top[x]],id[x],k);
		modify(1,id[top[x]],id[x],k);
		x=fa[top[x]];
	}
	if(dep[x]<dep[y])	swap(x,y);
	// printf("modify(1,%d,%d,%d)\n",id[y],id[x],k);
	modify(1,id[y],id[x],k);
}
int main(){
	n=RIN,m=RIN;
	foru(i,1,n-1){
		int u=RIN,v=RIN;
		add_e(u,v);
		add_e(v,u);
	}
	dfs1(1,0);
	dfs2(1,1);
	build(1,1,n);
	while(m--){
		char op;
		scanf("%c",&op);
		int x=RIN,y=RIN;
		if(op=='P'){
			updRange(x,y,1);
			int lca=LCA(x,y);
			updRange(lca,lca,-1);
		}else{
			printf("%d\n",qurey(1,id[(dep[x]>dep[y]?x:y)],id[(dep[x]>dep[y]?x:y)]));
		}
	}
	return 0;
}

RIN是快读,见上方define,a=RIN相当于a=read()

最后那个地方如果改成cin>>op>>x>>y就AC了

如果按代码里的写,那个测试点尽管只有一个Q,会输出很多次0

测试点如下:

in
7 7 
3 1 
3 2 
3 4 
3 5 
3 6 
3 7 
P 1 2 
P 4 2 
P 5 2 
P 6 2 
P 3 2 
P 7 2 
Q 3 2 

out
6
2022/8/3 19:15
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