HDU 2962 Trucking 二分+最短路

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题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2962

题意

在一张地图中，一个运货车要从s运货到t，路中每条边有运输时间，有些边有运货限制（理解为路有路有承重限制），有些没有。给定运货车的最大运货量，求s到t能运货尽可能多的情况下，花费时间最少。

思路

二分枚举s->t的路中最小的h，将图中所有小于h的边都删去，跑最短路。

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;#define INF2 9223372036854775807llconst int INF = ( 2e9 ) + 2;const ll maxn = 1010;const int maxm = maxn*maxn;struct edge{    int v,h,w,next;}e[maxm];struct node{    int u,d;    bool operator < (const node &b)const    {        return d>b.d;    }};int head[maxn],d[maxn],vis[maxn];int tot;void add(int u,int v,int h,int w){    e[tot].v=v;    e[tot].h=h;    e[tot].w=w;    e[tot].next=head[u];    head[u]=tot++;    e[tot].v=u;    e[tot].h=h;    e[tot].w=w;    e[tot].next=head[v];    head[v]=tot++;}int dijkstra(int s,int t,int n,int lim){    for(int i=1;i<=n;i++)d[i]=INF;    d[s]=0;    priority_queue<node> q;    memset(vis,0,sizeof(vis));    q.push(node{s,0});    while(!q.empty())    {        node temp=q.top();q.pop();        int u=temp.u;        if(vis[u])continue;        vis[u]=1;        for(int i=head[u];i!=-1;i=e[i].next)        {            if(e[i].h<lim)continue;            int v=e[i].v;            int w=e[i].w;            if(d[v]>d[u]+w)            {                d[v]=d[u]+w;                q.push(node{v,d[v]});            }        }    }    return d[t];}int main(){    int n,m;    int cas=0;//  freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&m)&&(n+m))    {        memset(head,-1,sizeof(head));        tot=0;        for(int i=0;i<m;i++)        {            int u,v,h,w;            scanf("%d%d%d%d",&u,&v,&h,&w);            if(h==-1)h=INF;            add(u,v,h,w);        }        int s,t,r;        scanf("%d%d%d",&s,&t,&r);        int l=0;        int ans=INF;        while(l<=r)        {            int mid=(l+r)>>1;            int temp=dijkstra(s,t,n,mid);            if(temp!=INF)            {                l=mid+1;                ans=temp;            }            else            r=mid-1;        }        if(cas!=0)puts("");        printf("Case %d:\n",++cas);        if(ans!=INF)        {            printf("maximum height = %d\n",r);            printf("length of shortest route = %d\n",ans);        }        else        printf("cannot reach destination\n");    }}

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