7.25航电练习赛 F题 trucking 解题报告(二分+Dij)

7.25航电练习赛 F题 trucking 解题报告(二分+Dij)

这道题实际上就是最短路，只不过增加了高度这个限制条件，由于时间限制是10s,我们不妨二分枚举出所有高度在1-limit之间的情况（时间复杂度是n^2log n，可以接受）取最大的即可。
PS:本人觉得此题最BT之处在于输出格式,Fuck！最后一个Case居然不要空行，害我PE十几次。

#include < iostream >
#include < cmath >
#include < cstdio >
#include < algorithm >
using   namespace  std;
#define  MAX 1001
#define  MAX_INTEGER  0x7fffffff


int  n,m;
int  graph[MAX][MAX];
int  copygraph[MAX][MAX];
int  graphheight[MAX][MAX];
bool  visit[MAX];
int  dis[MAX];
int  s,t,lim;
int  maxn;
int  shortestpath;
int  casenum;

bool  dijkstra( int  s, int  t, int  graph[MAX][MAX])
{
    memset(visit,false,sizeof(visit));
    int i;
    int j;
    
    
    for(i=1;i<=n;i++)
    {
        dis[i]=graph[s][i];
    }
    graph[s][s]=0;
    
    dis[s]=0;
    visit[s]=true;
    for(i=1;i<n;i++)
    {
        int mark_num=n;
        int mark_dis=MAX_INTEGER;
        int temp=MAX_INTEGER;
        for(j=1;j<=n;j++)
        {
            if(!visit[j]&&dis[j]<temp)
            {
                mark_num=j;
                temp=dis[j];
                
            }
        }
        if(temp==MAX_INTEGER)
            return false;
        visit[mark_num]=true;
        if(mark_num==t)
            return true;
        for(j=1;j<=n;j++)
        {
            if((!visit[j])&&(graph[mark_num][j]<MAX_INTEGER))
            {
                int newdis=dis[mark_num]+graph[mark_num][j];
                if(newdis<dis[j])
                {
                    dis[j]=newdis;
                }
            }
        }
    }
    return false;
    
}

int  main ()
{
//    FILE *in,*out;
//    in = fopen("trucking.in", "r");
//    out = fopen("out.txt","w");
      

    int i,j;
    casenum=0;
    while(scanf("%d%d",&n,&m))

    {
        casenum++;
        maxn=0;
        if(n==0&&m==0)
            break;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(i==j)
                    graph[i][j]=0;
                else    
                    graph[i][j]=MAX_INTEGER;
            }
        }
        for(i=1;i<=m;i++)
        {
            int a,b,height,len;
            scanf("%d%d%d%d",&a,&b,&height,&len);
            graph[a][b]=len;
            graph[b][a]=len;
            if(height==-1)

            {
                graphheight[a][b]=9999999;
                graphheight[b][a]=9999999;
            }
            else
            {
                graphheight[a][b]=height;
                graphheight[b][a]=height;
            }

        
        }
        scanf("%d%d%d",&s,&t,&lim);
        int l=1;
        int r=lim;
        while(l<=r)
        {

            int mid=(r+l)>>1;
            for(i=1;i<=n;i++)
            {

                for(j=1;j<=n;j++)
                {
                    copygraph[i][j]=graph[i][j];

                    if(graphheight[i][j]<mid&&i!=j)
                        copygraph[i][j]=MAX_INTEGER;
                }
            }
            if(dijkstra(s,t,copygraph)==false)
            {
                r=--mid;
                continue;
            }
            else
            {

                l=mid+1;
                maxn=mid;
                shortestpath=dis[t];
                continue;
            }
        }
        if(casenum>1)
            printf("\n");
        if(maxn==0)
            printf("Case %d:\ncannot reach destination\n",casenum);
        else
            printf("Case %d:\nmaximum height = %d\nlength of shortest route = %d\n",casenum,maxn,shortestpath);
    }
    return 0;
}