HDU-1839 Delay Constrained Maximum Capacity Path 最短路

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题意： n个点 m条路 T时间内最小容量最大

思路：二分最小容量 + 最短路

由于s数组定义只定义了n（ m条边）纠结了一下午找不出错 QAQ 还是太粗心

#include "stdio.h"#include "string.h"#include "vector"#include "queue"#include "algorithm"using namespace std;const int maxn = 10005;const int inf = 1<<29;int n,m,T,pos;int s[maxn*5];int dis[maxn];bool vis[maxn];struct node{    int v,cap,t,next;}edge[maxn*10];int head[maxn];void add( int a,int b,int c,int d ){    edge[++pos].v = b; edge[pos].cap = c; edge[pos].t = d;    edge[pos].next = head[a];    head[a] = pos;}typedef pair<int,int>pii;bool Dijkstra( int s,int Cap ){    for( int i = 1; i <= n; i ++ )        dis[i] = inf;    memset( vis,0,sizeof(vis) );    priority_queue<pii,vector<pii>,greater<pii> >que;    dis[s] = 0;    que.push( make_pair(dis[s],s) );    while( !que.empty() )    {        pii u = que.top(); que.pop();        int x = u.second;        vis[x] = true;        if( dis[x] > T ) return false;        if( x == n ) return dis[x];        for( int i = head[x]; i != -1; i = edge[i].next )        {            node v = edge[i];            if( v.cap < Cap || vis[v.v] ) continue;            if( dis[v.v] > dis[x] + v.t )            {                dis[v.v] = dis[x] + v.t;                que.push( make_pair(dis[v.v],v.v) );            }        }    }    return false;}void init(){    pos = 0;    for( int i = 1; i <= n; i ++ )        head[i] = -1;}int main(){    //freopen( "data.txt","r",stdin );    int t,a,b,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",&a,&b,&c,&d);            add(a,b,c,d);            add(b,a,c,d);            s[i] = c;        }        sort(s,s+m);        int ld = 0,rd = 0,mid;        for( int i = 1; i < m; i ++ )        {            if( s[i] != s[i-1] )                s[++rd] = s[i];        }        while( ld <= rd )        {            mid = (ld+rd)>>1;            if( Dijkstra( 1,s[mid] ) )            {                ld = mid+1;            }            else rd = mid-1;        }        printf("%d\n",s[rd]);    }    return 0;}