发现每个点的权值都可以表示成 \(\rm k\pm x\)。
那么对于新增的方程,\(\rm x_u+x_v=k\pm x/0\) 且 \(\rm x_u+x_v=s\)。
如果 \(x\) 项系数为 \(0\),那么就只需判断 \(\rm x_u+x_v=s\) 有无解。
若不为 \(0\),那么直接解出 \(x_1\) 并判断是否是小数即可。
修改操作就是对一段区间的值加或减,直接树状数组,复杂度 \(\mathcal O\rm((n+q)logn)\)
#include<bits/stdc++.h> #define ri register signed #define p(i) ++i namespace IO{ char buf[1<<21],*p1=buf,*p2=buf; #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++ struct nanfeng_stream{ template<typename T>inline nanfeng_stream operator>>(T &x) { ri f=0;x=0;register char ch=gc(); while(!isdigit(ch)) f|=ch=='-',ch=gc(); while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc(); return x=f?-x:x,*this; } }cin; } using IO::cin; namespace nanfeng{ #define FI FILE *IN #define FO FILE *OUT template<typename T>inline T cmax(T x,T y) {return x>y?x:y;} template<typename T>inline T cmin(T x,T y) {return x>y?y:x;} typedef long long ll; static const int N=1e6+7; int first[N],dep[N],ld[N],rd[N],ww[N],n,q,tot,t=1,opt,u,v; ll W[N],s; struct BIT{ #define lowbit(x) ((x)&-(x)) ll c[N]; inline void update(int x,ll k) {for (ri i(x);i<=n;i+=lowbit(i)) c[i]+=k;} inline ll query(int x) { ll res(0); for (ri i(x);i;i-=lowbit(i)) res+=c[i]; return res; } }B; struct edge{int v,w,nxt;}e[N]; inline void add(int u,int v,int w) {e[t].v=v,e[t].w=w,e[t].nxt=first[u],first[u]=t++;} void dfs(int x,ll w) { W[ld[x]=p(tot)]=w; for (ri i(first[x]),v;i;i=e[i].nxt) { dep[v=e[i].v]=dep[x]+1; if (dep[v]&1) dfs(v,w-e[i].w); else dfs(v,w+e[i].w); } rd[x]=tot; } inline int main() { //FI=freopen("nanfeng.in","r",stdin); //FO=freopen("nanfeng.out","w",stdout); cin >> n >> q; for (ri i(2),f;i<=n;p(i)) cin >> f >> ww[i],add(f,i,ww[i]); dfs(1,0); for (ri i(1),w;i<=tot;p(i)) w=W[i]-W[i-1],B.update(i,w); for (ri i(1);i<=q;p(i)) { cin >> opt; if (opt==2) { cin >> u >> s; if (dep[u]&1) B.update(ld[u],-s+ww[u]),B.update(rd[u]+1,s-ww[u]); else B.update(ld[u],s-ww[u]),B.update(rd[u]+1,-s+ww[u]); ww[u]=s; } else { cin >> u >> v >> s; ri jd(0); register ll tmp1=B.query(ld[u]),tmp2=B.query(ld[v]); if (dep[u]&1) --jd,tmp1*=-1ll;else p(jd); if (dep[v]&1) --jd,tmp2*=-1ll;else p(jd); if (!jd) { if (tmp1+tmp2!=s) puts("none"); else puts("inf"); } else if (jd==2) { register ll ans=(s-tmp1-tmp2)>>1ll; if ((ans<<1ll)+tmp1+tmp2==s) printf("%lld\n",ans); else puts("none"); } else { register ll ans=(tmp1+tmp2-s)>>1ll; if ((ans<<1ll)+s==tmp1+tmp2) printf("%lld\n",ans); else puts("none"); } } } return 0; } } int main() {return nanfeng::main();}