记录编号 |
583470 |
评测结果 |
AAAAAAAAAAAAAAAAAAAA |
题目名称 |
正则表达式 |
最终得分 |
100 |
用户昵称 |
┭┮﹏┭┮ |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
1.709 s |
提交时间 |
2023-10-15 15:59:20 |
内存使用 |
10.25 MiB |
显示代码纯文本
#include <bits/stdc++.h>
using namespace std;
//tarjan求强连通分量+缩点+最短路(spfa或dijkstra)
#define P pair<int,int>
#define make(x,y) make_pair(x,y)
const int N = 2e5+10,M = 1e6+10;
int n,m;
struct made{
int ver,nx,ed;
}e[M],re[M];//原图 以及 缩点重建图
int hd[N],rhd[N],tot,cnt,num,top;
int low[N],dfn[N],st[N],color[N];
bool v[N];//tarjan
map<P,int>mp;//判重边
void add(int x,int y,int z){
tot++;
e[tot].ver = y,e[tot].ed = z,e[tot].nx = hd[x],hd[x] = tot;
}//原图
int radd(int x,int y,int z){
tot++;
re[tot].ver = y,re[tot].ed = z,re[tot].nx = rhd[x],rhd[x] = tot;
return tot;
}//缩点图
void tarjan(int x) {
low[x] = dfn[x] = ++cnt;
st[++top] = x,v[x] = 1;
for(int i = hd[x];i;i = e[i].nx){
int y = e[i].ver;
if(!dfn[y])tarjan(y),low[x] = min(low[x],low[y]) ;
else if(v[y])low[x] = min(low[x],dfn[y]);
}
if(low[x] == dfn[x]){
num++;int y = 0;
do{
y = st[top--];
v[y] = 0,color[y] = num;
}while(x != y);
}
}//正常tarjan
void build(int x){
for(int i = hd[x];i;i = e[i].nx){
int y = e[i].ver;
int rx = color[x],ry = color[y];
int f = mp[make(rx,ry)];
if(rx == ry)continue;
if(f){
re[f].ed = min(re[f].ed,e[i].ed);//最优
continue;
}//判重边
mp[make(rx,ry)] = radd(rx,ry,e[i].ed);
}
}//重建缩点
int d[N];
priority_queue<P,vector<P>,greater<P> >q;
void dijkstra(int x){
memset(v,0,sizeof(v));
memset(d,0x3f,sizeof(d));
d[x] = 0;
q.push(make(d[x],x));
while(!q.empty()){
int x = q.top().second;q.pop();
if(v[x])continue;
v[x] = 1;
for(int i = rhd[x];i;i = re[i].nx){
int y = re[i].ver,z = re[i].ed;
if(d[y] > d[x] + z){
d[y] = d[x] + z;
q.push(make(d[y],y));
}
}
}
}//dijkstra求最短路
int main(){
freopen("regexp.in","r",stdin);
freopen("regexp.out","w",stdout);
scanf("%d%d",&n,&m);
for(int i = 1;i <= m;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
add(x,y,z);
}
for(int i = 1;i <= n;i++)
if(!color[i])tarjan(i);//
//
tot = 0;
for(int i = 1;i <= n;i++)
build(i);//
dijkstra(color[1]);//
printf("%d\n",d[color[n]]);
return 0;
}