hdu2962 Trucking (最短路+二分查找)

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

5 6

1 2 7 5

1 3 4 2

2 4 -1 10

2 5 2 4

3 4 10 1

4 5 8 5

1 5 10

5 6

1 2 7 5

1 3 4 2

2 4 -1 10

2 5 2 4

3 4 10 1

4 5 8 5

1 5 4

3 1

1 2 -1 100

1 3 10

0 0

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

这个题花了我和我老婆一下午的时间 本来以为很简单 后来越想越复杂 改了好多遍 最后还PE了一遍T^T，唉 。。。。结果还被我老婆说了

题目大意是有n个点m条路 每条路都有一个限高，最后给出一个起点 一个终点和车所能装货物的最大高度，求在车装货物高度尽量高的情况下的最短路

先用二分列举货物的高度 再用SPFA找最短路就OK0.0，，，噢噢噢噢 最后要注意下输出格式 是在测试数据之间有空行

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

struct node
{
    int h,len;
} map[1010][1010];

int start,end,height,c;
int node[1010];
const int inf=9999999;

int Spfa(int high)
{
    for (int i=1;i<=c;i++)
    node[i]=inf;
    queue<int>q;
    int inq[1010]= {0};
    int tm=start;
    node[tm]=0;
    inq[tm]=1;
    q.push(tm);
    while (!q.empty())
    {
        int s=q.front();
        q.pop();
        for (int i=1; i<=c; i++)
        {
            //cout<<s<<i<<" "<<node[i]<<" "<<map[s][i].len<<endl;
            if (map[s][i].h>=high&&node[i]>map[s][i].len+node[s])
            {
                node[i]=map[s][i].len+node[s];
                //cout<<"   "<<i<<" "<<node[i]<<endl;
                if (!inq[i])
                {
                    q.push(i);
                    inq[i]=1;
                }
            }
        }
        inq[s]=0;

    }
    if (node[end]!=inf)
        return node[end];
    else
        return -1;
}

int main ()
{
    int r,maxx,minn,h,k=1;
    while (cin>>c>>r&&(c||r))
    {
        int ans=-1,cmp=-1;
        for(int i=1;i<=c;i++)
        {
            for(int j=1;j<=c;j++)
            {
                map[i][j].len=inf;
                map[i][j].h=0;
            }
        }
        maxx=0,minn=inf;
        for (int i=1; i<=r; i++)
        {
            int a,b,len;

            cin>>a>>b>>h>>len;
            if(h==-1) h=inf;
            if (minn>h) minn=h;
            if (maxx<h) maxx=h;
            //cout<<minn<<" "<<maxx<<endl;
            if (map[a][b].len>len)
                map[a][b].len=map[b][a].len=len;
            if (map[a][b].h<h)
                map[a][b].h=map[b][a].h=h;
        }
        cin>>start>>end>>height;
        maxx=height>maxx?maxx:height;
        int l=minn,r=maxx;
        while (l<=r)
        {
            int mid=(r+l)>>1;
            //cout<<l<<" "<<r<<" "<<mid<<endl;
            int flag=Spfa(mid);
            if (flag!=-1)
            {
                l=mid+1;
                ans=mid;
                cmp=flag;
            }
            else
                r=mid-1;
        }
        if(k>1)
        printf("\n");
        printf("Case %d:\n",k++);
        if(ans==-1)
        printf("cannot reach destination\n");
        else
        {
            printf ("maximum height = %d\n",ans);
        printf ("length of shortest route = %d\n",cmp);
        }
    }
    return 0;
}

hdu2962 Trucking (最短路+二分查找)