poj 2263 Heavy Cargo 求a到b路上最小值中的最大值

最新推荐文章于 2021-05-24 00:50:20 发布

原创最新推荐文章于 2021-05-24 00:50:20 发布 · 507 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #each #path #up

acm_图论专栏收录该内容

71 篇文章

订阅专栏

本文介绍了一种用于确定两点间最大载重路径的算法。该算法适用于大型卡车制造商BigJohnsson Trucks Inc.的新车型Godzilla V12，确保在不同城市间的运输过程中不会违反道路的重量限制规定。通过输入城市的数量、道路段及其重量限制，算法能够计算出从起点到终点的最大可能负载。

Description

Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount of cargo you can transport with it is never limited by the truck itself. It is only limited by the weight restrictions that apply for the roads along the path you want to drive.

Given start and destination city, your job is to determine the maximum load of the Godzilla V12 so that there still exists a path between the two specified cities.

Input

The input will contain one or more test cases. The first line of each test case will contain two integers: the number of cities n (2<=n<=200) and the number of road segments r (1<=r<=19900) making up the street network.
Then r lines will follow, each one describing one road segment by naming the two cities connected by the segment and giving the weight limit for trucks that use this segment. Names are not longer than 30 characters and do not contain white-space characters. Weight limits are integers in the range 0 - 10000. Roads can always be travelled in both directions.
The last line of the test case contains two city names: start and destination.
Input will be terminated by two values of 0 for n and r.

Output

For each test case, print three lines:

a line saying "Scenario #x" where x is the number of the test case
a line saying "y tons" where y is the maximum possible load
a blank line

Sample Input

4 3
Karlsruhe Stuttgart 100
Stuttgart Ulm 80
Ulm Muenchen 120
Karlsruhe Muenchen
5 5
Karlsruhe Stuttgart 100
Stuttgart Ulm 80
Ulm Muenchen 120
Karlsruhe Hamburg 220
Hamburg Muenchen 170
Muenchen Karlsruhe
0 0

Sample Output

Scenario #1
80 tons
 
Scenario #2
170 tons

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
const int inf=(1<<28);
map<string,int>q;
int tq;
char str1[1000],str2[1000];
int a[300][300];
int n,e;
int main()
{
    int pl=1;
    while(scanf("%d%d",&n,&e)==2&&n)
    {
        q.clear();tq=0;
        memset(a,0,sizeof(a));
        for(int i=0;i<e;i++)
        {
            int w;
            scanf("%s%s%d",str1,str2,&w);
            if(q[str1]==0) q[str1]=++tq;
            if(q[str2]==0) q[str2]=++tq;
            a[q[str1]][q[str2]]=a[q[str2]][q[str1]]=w;
        }
        for(int k=1;k<=n;k++)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    a[i][j]=max(a[i][j],min(a[i][k],a[k][j]));
                }
            }
        }
        scanf("%s%s",str1,str2);
        printf("Scenario #%d\n%d tons\n\n",pl++,a[q[str1]][q[str2]]);
    }
    return 0;
}