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。
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#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,greater >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; } |
HDU 2962 Trucking,布布扣,bubuko.com
原文：http://www.cnblogs.com/forever97/p/3619187.html