题目链接：

https://www.luogu.org/problemnew/show/P1462

思路：

又是一道水题，很明显二分+最短路

而且这道题数据非常水，spfa有个小错误居然拿了91分还比正解快

我们二分金钱数，节点权值大于二分值的都不能走。二分中跑spfa，如果不能走到终点，即dis[n]>=b，则说明二分值偏小。

注意：

代码：

#include
#include
#include
#include
#include
#include
#include
#include
#define ri register int
#define ll long long
using namespace std;
const ll inf=9926081792;
const int maxn=10005;
const int maxm=50005;
ll g[maxn],dis[maxn];
int h[maxn],num_edge=0;
bool vis[maxn];
int n,m;
ll b;
struct Edge{
int ne,to;
ll dis;
}edge[maxm<<2]; inline void add_edge(int f,int t,ll dis){ edge[++num_edge].ne=h[f]; edge[num_edge].to=t; edge[num_edge].dis=dis; h[f]=num_edge; } template inline void read(T &x){
x=0;int ne=0;char c;
while(!isdigit(c=getchar()))ne=c=='-';
x=c-48;
while(isdigit(c=getchar()))x=(x<<3)+(x<<1)+c-48; x=ne?-x:x; return ; } inline int spfa(ll len){ dequeq;
int u,v,c;
while(q.size())q.pop_front();
//memset(vis,0,sizeof(vis));
for(ri i=1;i<=n;i++){dis[i]=inf,vis[i]=0;} dis[1]=0,vis[1]=1; q.push_back(1); while(!q.empty()){ u=q.front();q.pop_front(); for(ri i=h[u];i;i=edge[i].ne){ v=edge[i].to; if((len-g[v]>=0)&&dis[v]>dis[u]+edge[i].dis){
dis[v]=dis[u]+edge[i].dis;
if(q.empty()){
if(!vis[v]){
vis[v]=1;
q.push_back(v);
}
}
c=q.front();
if(dis[v]>dis[c]){
if(!vis[v]){
vis[v]=1;
q.push_back(v);
}
}
else if(!vis[v]){
vis[v]=1;
q.push_front(v);
}

        }
    }
    vis[u]=0;
}
if(dis[n]>=b)return 0;
return 1;

}
int main(){
ll l=1,r=n;
int u,v,c;
read(n),read(m),read(b);
for(ri i=1;i<=n;i++)read(g[i]); for(ri i=1;i<=m;i++){ read(u),read(v),read(c); add_edge(u,v,c); add_edge(v,u,c); } if(!spfa(inf)){ puts("AFK"); return 0; } while(l>1;
if(spfa(mid))r=mid;
else l=mid+1;
}
printf("%d",r);
return 0;
}