题目链接
题解
转移的代价是存在于边和边之间的
所以把边看做点，跑最短路
但是这样做需要把同一个点的所有入边和所有出边之间连边
\(O(m^2)\)的连边无法接受
需要优化建图
膜一下Claris的方法
对每个点，取出其入边出边，按在字典树上的\(dfs\)序排序
按\(dfs\)序排序，实际上就是将字符串排序了
按照后缀数组的理论，两点之间的\(lcp\)就是两点之间相邻\(lcp\)，也就是\(height\)数组的最小值
对于一个位置的\(height\)，两边的点之间联通所需最小代价不超过\(height\)
所以可以用\(lca\)求出\(height\)数组，建立前缀后缀虚点
以前缀为例，横向边为\(0\)，纵向边为\(height\)的大小
这样如果想从左边的点到达右边的点，就只需经过中间最小的\(height\)了
类似可以建立后缀点处理从右到左的情况
复杂度\(O(mlogm)\)
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 500005,maxm = 2000005,INF = 2000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
struct EDGE{int to,nxt,w;}ed[maxm];
struct Trie{
int h[maxn],ne,dfn[maxn],fa[maxn][16],dep[maxn],cnt,n;
EDGE ed[maxm];
void init(){REP(i,n) h[i] = 0; ne = 1; cnt = 0;}
void build(int u,int v){ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;}
void dfs(int u){
dfn[u] = ++cnt;
REP(i,15) fa[u][i] = fa[fa[u][i - 1]][i - 1];
Redge(u) {
fa[to = ed[k].to][0] = u;
dep[to] = dep[u] + 1; dfs(to);
}
}
int lca(int u,int v){
if (dep[u] < dep[v]) swap(u,v);
for (int D = dep[u] - dep[v],i = 0; (1 << i) <= D; i++)
if (D & (1 << i)) u = fa[u][i];
if (u == v) return u;
for (int i = 15; ~i; i--)
if (fa[u][i] != fa[v][i]){
u = fa[u][i];
v = fa[v][i];
}
return fa[u][0];
}
}T;
int h[maxn],ne = 1;
int n,m,N,w[maxn],pos[maxn];
int c[maxn],ci,height[maxn];
int pu[maxn],pd[maxn],su[maxn],sd[maxn];
int d[maxn],vis[maxn];
vector<int> in[maxn],out[maxn];
priority_queue<cp,vector<cp>,greater<cp> > q;
inline void build(int u,int v,int w){ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;}
inline bool cmp(const int& a,const int& b){
return T.dfn[pos[abs(a)]] < T.dfn[pos[abs(b)]];
}
void init(){
T.init(); ne = 1; N = 0; cls(h,0); cls(w,0);
REP(i,n) in[i].clear(),out[i].clear();
}
void Build(){
for (int p = 1; p <= n; p++){
if (!in[p].size() && !out[p].size()) continue;
ci = 0;
for (unsigned int i = 0; i < in[p].size(); i++)
c[++ci] = in[p][i];
for (unsigned int i = 0; i < out[p].size(); i++)
c[++ci] = -out[p][i];
sort(c + 1,c + 1 + ci,cmp);
for (int i = 1; i <= ci; i++){
pu[i] = ++N,pd[i] = ++N;
su[i] = ++N,sd[i] = ++N;
if (i > 1){
build(pu[i - 1],pu[i],0);
build(pd[i - 1],pd[i],0);
build(su[i],su[i - 1],0);
build(sd[i],sd[i - 1],0);
}
if (c[i] >= 0) build(c[i],pu[i],0),build(c[i],su[i],0);
else c[i] *= -1,build(pd[i],c[i],0),build(sd[i],c[i],0);
}
for (int i = 1; i < ci; i++){
int tmp = T.dep[T.lca(pos[c[i]],pos[c[i + 1]])];
build(pu[i],pd[i + 1],tmp);
build(su[i + 1],sd[i],tmp);
}
}
}
void dijkstra(){
for (int i = 0; i <= N; i++) d[i] = INF,vis[i] = false;
for (unsigned int j = 0; j < out[1].size(); j++)
q.push(mp(d[out[1][j]] = 0,out[1][j]));
int u;
while (!q.empty()){
u = q.top().second; q.pop();
if (vis[u]) continue;
vis[u] = true;
Redge(u) if (!vis[to = ed[k].to] && d[to] > d[u] + ed[k].w + w[u]){
d[to] = d[u] + ed[k].w + w[u];
q.push(mp(d[to],to));
}
}
}
void print(){
for (int i = 2; i <= n; i++){
int ans = INF;
for (unsigned int j = 0; j < in[i].size(); j++)
ans = min(ans,d[in[i][j]] + w[in[i][j]]);
printf("%d\n",ans);
}
}
int main(){
int Case = read();
while (Case--){
n = read(); m = read(); T.n = read(); init();
int a,b;
REP(i,m){
a = read(); b = read();
w[i] = read(); pos[i] = read();
out[a].push_back(i);
in[b].push_back(i);
}
N = m;
for (int i = 1; i < T.n; i++){
a = read(); b = read(); read();
T.build(a,b);
}
T.dfs(1);
Build();
dijkstra();
print();
}
return 0;
}
转播
