Roads in the Kingdom CodeForces - 835F (直径)
作者:互联网
大意: 给定一个基环树, 求删除一条环上的边使得直径最小.
直径分两种情况
- 环上点延伸的树内的直径
- 两个环上点的树内深度最大的点匹配
第一种情况直接树形dp求一下, 第二种情况枚举删除的环边, 线段树维护一下即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n,dep[N],vis[N]; struct _ {int to,w;} fa[N]; vector<_> g[N]; int a[N], v[N], top; ll mx[N], b[N], sum[N], ans, ans2; struct node { ll m1,m2,v; node operator + (const node &rhs) const { node ret; ret.m1=max(m1,rhs.m1); ret.m2=max(m2,rhs.m2); ret.v=max(v,rhs.v); ret.v=max(ret.v,m2+rhs.m1); return ret; } } tr[N<<2]; void get(int x, int y) { if (dep[x]<dep[y]) return; v[++top] = y; a[top+1] = fa[y].w; for (; x!=y; x=fa[x].to) v[++top] = x, a[top+1] = fa[x].w; } void dfs(int x, int f) { dep[x] = dep[f]+1; for (_ e:g[x]) if (e.to!=f) { int y = e.to; fa[y] = {x,e.w}; if (dep[y]) get(x,y); else dfs(y,x); } } void dfs2(int x) { vis[x] = 1; for (_ e:g[x]) if (!vis[e.to]) { int y = e.to; dfs2(y); ans = max(ans, mx[x]+mx[y]+e.w); mx[x] = max(mx[x], mx[y]+e.w); } } void build(int o, int l, int r) { if (l==r) tr[o].m1=b[l]+sum[l],tr[o].m2=b[l]-sum[l],tr[o].v=0; else { build(ls),build(rs); tr[o]=tr[lc]+tr[rc]; } } node qry(int o, int l, int r, int ql, int qr) { if (ql<=l&&r<=qr) return tr[o]; if (mid>=qr) return qry(ls,ql,qr); if (mid<ql) return qry(rs,ql,qr); return qry(ls,ql,qr)+qry(rs,ql,qr); } int main() { scanf("%d", &n); REP(i,1,n) { int u, v, w; scanf("%d%d%d",&u,&v,&w); g[u].pb({v,w}); g[v].pb({u,w}); } dfs(1,0); REP(i,1,top) vis[v[i]]=1; REP(i,1,top) dfs2(v[i]),b[i]=mx[v[i]]; REP(i,top+1,2*top-1) b[i] = b[i-top]; REP(i,top+2,2*top-1) a[i] = a[i-top]; REP(i,1,2*top-1) sum[i] = sum[i-1]+a[i]; build(1,1,2*top-1); ans2 = 1e18; REP(i,1,top) ans2 = min(ans2, qry(1,1,2*top-1,i,i+top-1).v); printf("%lld\n",max(ans,ans2)); }
标签:Kingdom,return,ll,CodeForces,ret,m1,Roads,include,define 来源: https://www.cnblogs.com/uid001/p/11623580.html