Bzoj4016: [FJOI2014]最短路径树问题

题面

传送门

Sol

先\(SPFA\)求出单源最短路，\(Bfs\)建出树，字典序可以用堆解决
然后就是点分治的一眼题
开桶记录到当前根经过边长度相同的最长路，记录它的长度

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自己强行\(yy\)了一个这种类型的点分丑陋写法

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);IL int Input(){RG int x = 0, z = 1; RG char c = getchar();for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);return x * z;
}int n, m, k, mx[_], root, size[_], vis[_], tot, dis[_];
struct Edge{int to[_], w[_], next[_], first[_], cnt;IL void Init(){Fill(first, -1);}IL void Add(RG int u, RG int v, RG int ww){to[cnt] = v, next[cnt] = first[u], w[cnt] = ww, first[u] = cnt++;}
} G1, G2;
int deep[_], len[_], t[_], num[_], S1[_], S2[_];
ll ans1, ans2;
queue <int> Q;
priority_queue <int> H;IL void GetRoot(RG int u, RG int ff){size[u] = 1, mx[u] = 0;for(RG int e = G2.first[u]; e != -1; e = G2.next[e]){RG int v = G2.to[e];if(vis[v] || v == ff) continue;GetRoot(v, u);size[u] += size[v];mx[u] = max(mx[u], size[v]);}mx[u] = max(mx[u], tot - size[u]);if(mx[u] < mx[root]) root = u;
}IL void GetDeep(RG int u, RG int ff, RG int dd, RG int nn){if(nn > k) return;S1[++S1[0]] = u, S2[++S2[0]] = u, deep[u] = dd, len[u] = nn;for(RG int e = G2.first[u]; e != -1; e = G2.next[e]){RG int v = G2.to[e], w = G2.w[e];if(vis[v] || v == ff) continue;GetDeep(v, u, dd + w, nn + 1);}
}IL void Solve(RG int u){vis[u] = 1, t[0] = 0, num[0] = 1;for(RG int e = G2.first[u]; e != -1; e = G2.next[e]){RG int v = G2.to[e];if(vis[v]) continue;GetDeep(v, u, G2.w[e], 1);for(RG int i = 1; i <= S1[0]; ++i){RG int dd = deep[S1[i]], nn = k - len[S1[i]];if(!num[nn]) continue;if(t[nn] + dd > ans1) ans1 = t[nn] + dd, ans2 = num[nn];else if(t[nn] + dd == ans1) ans2 += num[nn];}for(; S1[0]; --S1[0]){RG int dd = deep[S1[S1[0]]], nn = len[S1[S1[0]]];if(dd > t[nn]) t[nn] = dd, num[nn] = 1;else if(dd == t[nn]) num[nn]++;}}for(; S2[0]; --S2[0]){RG int dd = deep[S2[S2[0]]], nn = len[S2[S2[0]]];t[nn] = num[nn] = 0;}for(RG int e = G2.first[u]; e != -1; e = G2.next[e]){RG int v = G2.to[e];if(vis[v]) continue;root = 0, tot = size[v];GetRoot(v, u), Solve(root);}
}IL void SPFA(){Fill(dis, 127);Q.push(1), vis[1] = 1, dis[1] = 0;while(!Q.empty()){RG int u = Q.front(); Q.pop();for(RG int e = G1.first[u]; e != -1; e = G1.next[e]){RG int v = G1.to[e], w = G1.w[e];if(dis[u] + w < dis[v]){dis[v] = dis[u] + w;if(!vis[v]) vis[v] = 1, Q.push(v);}}vis[u] = 0;}
}IL void Build(){H.push(-1), Fill(vis, 0), vis[1] = 1;while(!H.empty()){RG int u = -H.top(); H.pop();for(RG int e = G1.first[u]; e != -1; e = G1.next[e]){RG int v = G1.to[e], w = G1.w[e];if(vis[v] || dis[v] != dis[u] + w) continue;G2.Add(u, v, w), G2.Add(v, u, w);vis[v] = 1, H.push(-v);}}
}int main(RG int argc, RG char* argv[]){tot = n = Input(), m = Input(), k = Input() - 1;G1.Init(), G2.Init();for(RG int i = 1; i <= m; ++i){RG int u = Input(), v = Input(), w = Input();G1.Add(u, v, w), G1.Add(v, u, w);}SPFA(), Build(), Fill(vis, 0);mx[0] = n + 1, GetRoot(1, 0),Solve(root);printf("%lld %lld\n", ans1, ans2);return 0;
}