题面
Sol
巧妙的建图+(Dijkstra)
考虑把边看成点，那么显然暴力建图的边数是(m^2)的
考虑优化
把(max(a, b))变成(a+max(b-a,0))
把每个点连出的边按权值从小到大排序
每个边向后面的边连(b-a), 后面向前面连(0)
连向它的边向连出去的边连(a)
新建(S)点，向(1)连出的边连，新建(T)，连向(n)的边连(T)
没开long long WA一片
# 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 _(4e5 + 5);
typedef int Arr[_];
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;
}
Arr first, vis;
ll dis[_];
int n, m, cnt, S, T;
struct Data{
int u; ll dis;
IL int operator <(RG Data B) const{
return dis > B.dis;
}
};
priority_queue <Data> Q;
struct Edge{
int to, next, w;
} edge[_ << 2];
struct Link{
int v, w, id;
IL int operator <(RG Link B) const{
return w < B.w;
}
};
vector <Link> G[_];
IL void Add(RG int u, RG int v, RG int w){
edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}
IL void Dijkstra(){
Fill(dis, 63), dis[S] = 0, Q.push((Data){S, 0});
while(!Q.empty()){
RG Data x = Q.top(); Q.pop();
if(vis[x.u]) continue;
vis[x.u] = 1;
for(RG int e = first[x.u]; e != -1; e = edge[e].next){
RG int v = edge[e].to, w = edge[e].w;
if(dis[x.u] + w < dis[v]) Q.push((Data){v, dis[v] = dis[x.u] + w});
}
}
}
int main(RG int argc, RG char *argv[]){
Fill(first, -1), n = Input(), m = Input();
for(RG int i = 1; i <= m; ++i){
RG int u = Input(), v = Input(), w = Input();
G[u].push_back((Link){v, w, i});
G[v].push_back((Link){u, w, m + i});
}
T = m + m + 1; RG int tmp = m;
for(RG int i = 1; i <= n; ++i){
sort(G[i].begin(), G[i].end()); RG int l = G[i].size();
for(RG int j = 0; j < l; ++j){
Add((G[i][j].id > tmp ? G[i][j].id - m : G[i][j].id + m), G[i][j].id, G[i][j].w);
if(i == 1) Add(S, G[i][j].id, G[i][j].w);
if(G[i][j].v == n) Add(G[i][j].id, T, G[i][j].w);
}
for(RG int j = 0; j < l - 1; ++j){
Add(G[i][j].id, G[i][j + 1].id, G[i][j + 1].w - G[i][j].w);
Add(G[i][j + 1].id, G[i][j].id, 0);
}
}
Dijkstra();
printf("%lld
", dis[T]);
return 0;
}