题目大意：给定无向图，每一条路上都有限重，求能到达目的地的最大限重，同时算出其最短路。
题解：由于有限重，所以二分检索，将二分的值代入最短路中，不断保存和更新即可。
#include <cstdio>
#include <cstring>
#include <utility>
#include <queue>
using namespace std;
const int N=20005;
const int INF=9999999;
typedef pair<int,int>seg;
priority_queue<seg,vector<seg>,greater<seg> >q;
int l,r,mid,begin,end,d[N],head[N],u[N],v[N],w[N],next[N],le[N],n,m,a,b,c;
int height,route;
bool vis[N];
void build(){
memset(head,-1,sizeof(head));
for(int e=1;e<=m;e++){
scanf("%d%d%d%d",&u[e],&v[e],&le[e],&w[e]);
if(le[e]==-1)le[e]=INF;
u[e+m]=v[e]; v[e+m]=u[e]; w[e+m]=w[e]; le[e+m]=le[e];
next[e]=head[u[e]]; head[u[e]]=e;
next[e+m]=head[u[e+m]]; head[u[e+m]]=e+m;
}
}
void Dijkstra(int src,int limit){
memset(vis,0,sizeof(vis));
for(int i=0;i<=n;i++) d[i]=INF;
d[src]=0;
q.push(make_pair(d[src],src));
while(!q.empty()){
seg now=q.top(); q.pop();
int x=now.second;
if(vis[x]) continue; vis[x]=true;
for(int e=head[x];e!=-1;e=next[e])
if(d[v[e]]>d[x]+w[e]&&le[e]>=limit){
d[v[e]]=d[x]+w[e];
q.push(make_pair(d[v[e]],v[e]));
}
}
}
int main(){
int cnt=1;
while(~scanf("%d%d",&n,&m)){
height=route=0;
if(m==0&&n==0)break;
if(cnt!=1) puts("");
printf("Case %d:\n",cnt++);
build();
scanf("%d%d%d",&begin,&end,&r);
l=1;
while(l<=r){
mid=(l+r)>>1;
Dijkstra(begin,mid);
if(d[end]!=INF){height=mid;route=d[end];l=mid+1;}
else r=mid-1;
}
if(height==0)puts("cannot reach destination");
else{
printf("maximum height = %d\n",height);
printf("length of shortest route = %d\n",route);
}
}
return 0;
}