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;

}