P1462 road to orgerma, p1462 road to orgerma

P1462 road to orgerma, p1462 road to orgerma
Background

On the continent of azelas, there is a magic named "grin". He is the backbone of the tribe.

One day, when he woke up, he found himself in the consortium's main city, stormwind.

After being attacked by a large number of consortium soldiers, he decided to escape to his hometown ogreema.

Description

There are n cities in azelas. 1, 2, 3,..., n.

There are m two-way roads between cities, connecting two cities. from one city to another, the city will be attacked by the Alliance, which will cause a loss of blood.

A certain toll (including the start and end points) will be collected each time you pass through a city ). There is no toll station on the road.

Assume that 1 is stormwind, n is ogreema, and his blood volume is B at most, his blood volume is full at departure.

He doesn't want to spend a lot of money. He wants to know what is the minimum charge for the most one of the cities he passes by when he can reach ogreima.

Input/Output Format Input Format:

The first row has three positive integers, n, m, and B. It indicates that there are n cities, m roads, and bits, respectively.

Next there are n rows, each row has 1 positive integer, fi. It indicates that the user passes through the city I and needs to pay the fi yuan.

Then there are m rows, with three positive integers in each row: ai, bi, and ci (1 <= ai, bi <= n ). It indicates that there is a highway between city ai and city bi. From city ai to city bi, or from city bi to city ai, the blood volume of ci will be lost.

Output Format:

Only one integer indicates the minimum value of the maximum payment.

If he cannot reach ogrema, output AFK.

Input and Output sample Input example #1:

4 4 8856102 1 22 4 11 3 43 4 3

Output sample #1:

10

Description

For 60% of data, n ≤ 200, m ≤ 10000, and B ≤ 200

For 100% of data, n ≤ 10000, m ≤ 50000, and B ≤ 1000000000

For 100% of the data, the ci ≤ 1000000000 and fi ≤ 1000000000 may have two sides connected to the same city.

 

Binary answer + SPFA

Be sure to enable long.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cmath>  5 #include<queue>  6 #define lli long long int   7 using namespace std;  8 void read(lli & n)  9 { 10     char c='+';lli x=0; 11     while(c<'0'||c>'9') 12     c=getchar(); 13     while(c>='0'&&c<='9') 14     { 15         x=x*10+(c-48); 16         c=getchar(); 17     } 18     n=x; 19 } 20 const lli MAXN=100001; 21 const lli maxn=0x7fffffff; 22 struct node 23 { 24     lli u,v,w,nxt; 25 }edge[MAXN]; 26 lli head[MAXN]; 27 lli num=1; 28 lli dis[MAXN]; 29 lli vis[MAXN]; 30 lli n,m,maxnblood; 31 lli spend[MAXN]; 32 lli l=0x7fffffff,r=-1; 33 void add_edge(lli x,lli y,lli z) 34 { 35     edge[num].u=x; 36     edge[num].v=y; 37     edge[num].w=z; 38     edge[num].nxt=head[x]; 39     head[x]=num++; 40 } 41 lli SPFA(lli r) 42 { 43     if(r<spend[1]) 44     {return 0;} 45     for(lli i=1;i<=n+1;i++) 46         dis[i]=maxn,vis[i]=0; 47     queue<int>q; 48     q.push(1); 49     vis[1]=1; 50     dis[1]=0; 51     while(q.size()!=0) 52     { 53         lli p=q.front(); 54         q.pop(); 55         vis[p]=0; 56         for(lli i=head[p];i!=-1;i=edge[i].nxt) 57         { 58             lli will=edge[i].v; 59             if((dis[will]>dis[p]+edge[i].w)&&spend[will]<=r) 60             { 61                 dis[will]=dis[p]+edge[i].w; 62                 if(vis[will]==0) 63                 { 64                     vis[will]=1; 65                     q.push(will); 66                 } 67             } 68         } 69     } 70     if(dis[n]<=maxnblood) 71     return 1; 72     else 73     return 0; 74 } 75 int main() 76 { 77     read(n);read(m);read(maxnblood); 78     for(lli i=1;i<=n;i++) 79         head[i]=-1; 80     for(lli i=1;i<=n;i++) 81     { 82         read(spend[i]); 83         l=min(spend[i],l); 84         r=max(spend[i],r); 85     } 86     for(lli i=1;i<=m;i++) 87     { 88         lli x,y,z; 89         read(x);read(y);read(z); 90         add_edge(x,y,z); 91         add_edge(y,x,z); 92     } 93     lli ans=0; 94     while(l<=r) 95     { 96         lli mid=(l+r)/2; 97         if(SPFA(mid)) 98         { 99             ans=mid;100             r=mid-1;101         }102         else l=mid+1;103     }104     //printf("%d",r);105     if(ans&&SPFA(ans))106     printf("%d",ans);107     else 108     printf("AFK");109     return 0;110 }