UVA-11374(最短路)

题意：

机场快线有经济线和商业线,现在分别给出经济线和商业线的的路线,现在只能坐一站商业线,其他坐经济线,问从起点到终点的最短用时是多少,还有路线是怎样的;

思路：

预处理出起点到所有站的最短距离和终点到所有站的最短距离,枚举要坐的那趟商业线,然后里面最小的就是答案了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e5+10;
const int maxn=500+10;
const double eps=1e-6;

int n,s,e,m;
int mp[maxn][maxn],dis[2][maxn],vis[maxn],p[2][maxn];

inline void Dijkstra(int from,int flag)
{
mst(vis,0);
vis[from]=1;
For(i,1,n)dis[flag][i]=mp[from][i],p[flag][i]=from;
dis[flag][from]=0;
For(i,2,n)
{
int temp=inf,k;
For(j,1,n)
if(!vis[j]&&temp>dis[flag][j])k=j,temp=dis[flag][j];
if(temp==inf)break;
vis[k]=1;
For(j,1,n)
{
if(!vis[j]&&dis[flag][j]>dis[flag][k]+mp[k][j])
dis[flag][j]=dis[flag][k]+mp[k][j],p[flag][j]=k;
}
}
}

int main()
{
int Case=0;
while(cin>>n>>s>>e)
{
if(Case)printf("\n");
Case++;
For(i,1,n)For(j,1,n)mp[i][j]=inf;
read(m);
int u,v,w;
For(i,1,m)
{
read(u);read(v);read(w);
mp[u][v]=mp[v][u]=min(mp[u][v],w);
}
Dijkstra(s,0);
Dijkstra(e,1);
int ans=dis[0][e],uppos=-1,downpos=-1;
int q;
read(q);
For(i,1,q)
{
read(u);read(v);read(w);
int temp=dis[0][u]+w+dis[1][v];
// cout<<dis[0][u]<<" "<<dis[1][v]<<endl;
if(temp<ans)ans=temp,uppos=u,downpos=v;
temp=dis[0][v]+w+dis[1][u];
if(temp<ans)ans=temp,uppos=v,downpos=u;
}
if(uppos==-1)
{
int cur=s;
while(cur!=e)printf("%d ",cur),cur=p[1][cur];
printf("%d\nTicket Not Used\n%d\n",e,ans);
}
else
{
int cur=uppos,cnt=0,tion[maxn];
while(cur!=s)
{
tion[++cnt]=cur;
cur=p[0][cur];
}
tion[++cnt]=s;
for(int i=cnt;i>0;i--)printf("%d ",tion[i]);
cur=downpos;
while(cur!=e)printf("%d ",cur),cur=p[1][cur];
printf("%d\n%d\n%d\n",e,uppos,ans);
}
//printf("\n");
}
return 0;
}