Delay Constrained Maximum Capacity Path HDU

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题目要求 首先要按时到达 其次尽可能走最短路线

通过二分 看能否在以当前长度的边为最短边的情况下按时到达 若不能则下调 否则上调

因为一条路径中最短边越长 对这条路径中边的要求也就越高 某些费时更少的边就被排除了 可能导致无法按时到达

反之 超过时限的可能也就越小

#include <stdio.h>#include <vector>#include <queue>#include <algorithm>#define N 99999999using namespace std;struct node1{    int v;    long long cap;    int con;};struct node2{    friend bool operator < (node2 n1,node2 n2)    {        return n1.s>n2.s;    }    int id;    int s;};vector <struct node1> edge[10001];long long len[50001];long long ans;int dis[10001],book[10001];int n,m,time,limit;void binsearch();int calculate();int main(){    struct node1 t;    long long c;    int cnt,i,j,a,b,d;    scanf("%d",&cnt);    while(cnt--)    {        scanf("%d%d%d",&n,&m,&time);        for(i=1;i<=n;i++)        {            edge[i].clear();        }        for(i=1;i<=m;i++)        {            scanf("%d%d%lld%d",&a,&b,&c,&d);            t.v=b,t.cap=c,t.con=d;            edge[a].push_back(t);            t.v=a,t.cap=c,t.con=d;            edge[b].push_back(t);            len[i]=c;        }        sort(len+1,len+m+1);        binsearch();        printf("%lld\n",ans);    }    return 0;}void binsearch(){    int l,mid,r;    l=1,r=m,ans=len[l];    while(l<=r)    {        mid=(l+r)/2;        limit=len[mid];        if(calculate())        {            ans=len[mid];            l=mid+1;        }        else        {            r=mid-1;        }    }    return;}int calculate(){    priority_queue <struct node2> que;    struct node1 t;    struct node2 cur,tem;    int i,p;    for(i=1;i<=n;i++)    {        dis[i]=N;        book[i]=0;    }    dis[1]=0;    tem.id=1;    tem.s=0;    que.push(tem);    while(!que.empty())    {        cur=que.top();        que.pop();        p=cur.id;        if(book[p]==1) continue;        book[p]=1;        for(i=0;i<edge[p].size();i++)        {            t=edge[p][i];            if(t.cap>=limit&&book[t.v]==0&&dis[t.v]>dis[p]+t.con)            {                dis[t.v]=dis[p]+t.con;                tem.id=t.v;                tem.s=dis[t.v];                que.push(tem);            }        }    }    if(dis[n]<=time) return 1;    else return 0;}