HDU1839(最短路+二分)

来源：互联网发布：美孚一号5w40数据编辑：程序博客网时间：2024/05/18 03:04

在最大高度下求最短路径，我们将高度排序，然后二分高度就行，因为节点比较多用邻接表存储

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int INF = 0x7fffffff;const int VN  =  10005;const int EN  =  500005;struct Edge{    int v,next,cap,time;} E[EN];int n, m, t;int size;int head[VN];int cap[EN];int d[VN];int Time[VN];int limit;bool inq[VN];void init(){    size=0;    memset(head, -1, sizeof(head));}void addEdge(int u,int v,int c,int d){    E[size].v=v;    E[size].cap=c;    E[size].time=d;    E[size].next = head[u];    head[u] = size++;}int Dijkstra(int src){    fill(d,d+n+1,INF);    d[src] = 0;    queue<int>q;    q.push(src);    while(!q.empty())    {        int u = q.front();        q.pop();        for(int e=head[u]; e!=-1; e=E[e].next)        {            if(E[e].cap>=limit)            {                int tmp = d[u]+E[e].time;                if(d[E[e].v] > tmp)                {                    d[E[e].v] = tmp;                    q.push(E[e].v);                }            }        }    }    return d[n];}int main(){    int T, u, v, c, d;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&t);        init();        for(int i=0; i<m; ++i)        {            scanf("%d%d%d%d",&u,&v,&c,&d);            cap[i]=c;            addEdge(u,v,c,d);            addEdge(v,u,c,d);        }        sort(cap, cap+m);        // 二分求解        int left=0, right=m-1, mid;        int ans = cap[0];        while(left <= right)        {            mid = (left+right)>>1;            limit = cap[mid];            int tmp=Dijkstra(1);            if(tmp==INF || tmp>t)            {                right = mid - 1;            }            else            {                ans = cap[mid], left = mid + 1;            }        }        printf("%d\n", ans);    }    return 0;}

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