Trucking

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 72 Accepted Submission(s): 28

Problem Description

A certain local trucking company would like to transport some goods on a cargo truck from one place to another. It is desirable to transport as much goods as possible each trip. Unfortunately, one cannot always use the roads in the shortest route: some roads may have obstacles (e.g. bridge overpass, tunnels) which limit heights of the goods transported. Therefore, the company would like to transport as much as possible each trip, and then choose the shortest route that can be used to transport that amount.

For the given cargo truck, maximizing the height of the goods transported is equivalent to maximizing the amount of goods transported. For safety reasons, there is a certain height limit for the cargo truck which cannot be exceeded.

Input

The input consists of a number of cases. Each case starts with two integers, separated by a space, on a line. These two integers are the number of cities (C) and the number of roads (R). There are at most 1000 cities, numbered from 1. This is followed by R lines each containing the city numbers of the cities connected by that road, the maximum height allowed on that road, and the length of that road. The maximum height for each road is a positive integer, except that a height of -1 indicates that there is no height limit on that road. The length of each road is a positive integer at most 1000. Every road can be travelled in both directions, and there is at most one road connecting each distinct pair of cities. Finally, the last line of each case consists of the start and end city numbers, as well as the height limit (a positive integer) of the cargo truck. The input terminates when C = R = 0.

Output

For each case, print the case number followed by the maximum height of the cargo truck allowed and the length of the shortest route. Use the format as shown in the sample output. If it is not possible to reach the end city from the start city, print "cannot reach destination" after the case number. Print a blank line between the output of the cases.

Sample Input

Sample Output

Case 1:
maximum height = 7
length of shortest route = 20

Case 2:
maximum height = 4
length of shortest route = 8

Case 3:
cannot reach destination

思路：题意：每一个地方有限高h，距离，车的最大载重量就是经过的地方限高的最小值，求这个最小值最大是多少，已经经过的路程

我们先二分这个h，判断mid是否合法，那么这样就可以做最短路时，做最短路时，判断mid是否小于限高即可啦，如果最后d[end]是INF，那么就不行，否则就行，

注意二分的边界！

  1 #include <cstdio>
  2 #include <cstring>   
  3 #include <iostream>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <queue>
  8 using namespace std;
  9 
 10 const int maxn=1010,maxm=20010,INF=10000000;
 11 struct qq
 12 {
 13     int n,to,z,h,ne;
 14     friend bool operator < (qq a,qq b)
 15     {
 16         return a.z>b.z;
 17     }
 18 } e[maxm],s,ya;
 19 
 20 priority_queue<qq> q;
 21 int d[maxn],node,x,y,z,cnt,to,n,m,h[maxn];
 22 bool f[maxn];
 23 int ans,st,en,height,lim,cas,minn,maxx,mid;
 24 
 25 void addedge(int x,int y,int z,int height)
 26 {
 27     cnt++;
 28     e[cnt].n=x;
 29     e[cnt].to=y;
 30     e[cnt].z=z;
 31     e[cnt].ne=h[x];
 32     e[cnt].h=height;
 33     h[x]=cnt;
 34 }
 35 
 36 void close()
 37 {
 38     exit(0);
 39 }
 40 
 41 void dijkstra(int st,int k)
 42 {
 43     memset(f,false,sizeof(f));
 44     while (!q.empty())
 45         q.pop();
 46     f[st]=true;
 47     for (int i=0;i<=n;i++)
 48         d[i]=INF;
 49     d[st]=0;
 50     for (int p=h[st];p!=-1;p=e[p].ne)
 51     {
 52         s.n=st;
 53         s.to=e[p].to;
 54         s.z=e[p].z;
 55         s.h=e[p].h;
 56         if (s.h<k && s.h!=-1)//高度不够
 57             continue;
 58         q.push(s);
 59     }
 60     while (!q.empty())
 61     {
 62         s=q.top();
 63         q.pop();
 64         to=s.to;
 65         if (f[to]) continue;
 66         d[to]=s.z;
 67         f[to]=true;
 68         for (int p=h[to];p!=-1;p=e[p].ne)
 69         {
 70             node=e[p].to;
 71             if (not f[node])
 72             {
 73                 ya.n=to;
 74                 ya.to=node;
 75                 ya.z=d[to]+e[p].z;
 76                 ya.h=e[p].h;
 77                 if (ya.h<k && ya.h!=-1) //高度不够
 78                     continue;
 79                 q.push(ya);
 80             }
 81         }
 82     }
 83 }
 84 
 85 bool judge(int mid)
 86 {
 87     dijkstra(st,mid);
 88     if (d[en]!=INF)
 89         return true;
 90     return false;
 91 }
 92 
 93 void work()
 94 {
 95     if (not judge(1))
 96     {
 97         printf("cannot reach destination\n");
 98         return;
 99     }
100     minn=1;maxx=lim;
101     while (maxx>minn)
102     {
103         mid=(maxx+minn+1)>>1;
104         if (judge(mid)) // The height is accepted!
105             minn=mid;
106         else
107             maxx=mid-1;
108     }
109     judge(maxx);
110     printf("maximum height = %d\n",maxx);
111     printf("length of shortest route = %d\n",d[en]);
112 }
113 
114 void init()
115 {
116 cas=0;
117     while (scanf("%d %d",&n,&m)!=EOF)
118     {
119         if (n==0 && m==0) break;
120         cas++;
121         if (cas!=1)
122             printf("\nCase %d:\n",cas);
123         else
124             printf("Case %d:\n",cas);
125         memset(h,-1,sizeof(h));
126         cnt=0;
127         for (int i=1;i<=m;i++)
128         {
129             scanf("%d %d %d %d",&x,&y,&z,&height);
130             addedge(x,y,height,z);
131             addedge(y,x,height,z);
132         }
133         scanf("%d %d %d",&st,&en,&lim);
134         work();
135     }
136 }
137 
138 
139 int main ()
140 {
141     init();
142     close();
143 }

posted on 2013-04-26 22:51 cssystem 阅读(178) 评论(0) 编辑收藏举报