Heavy Cargo（数论）

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

解题思路

这题和之前做的一道最短路很是相像，Floyd算法

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int map[205][205], num_space = 1;
char a[50], b[50], name[205][50];                                //name[]存放地址名

int find(char *a)                                                //地点与下标映射
{
    for(int i = 1; i < 201; i++)
        if(strcmp(a, name[i]) == 0)
            return i;
    int len = strlen(a);
    strcpy( name[num_space], a );
    name[num_space][len] = '\0';
    num_space++;
    return num_space - 1;
}

int main()
{
    int n, m, weight, cases = 1;
    while( scanf("%d%d", &n, &m) && (n || m) )
    {
        memset(map, -1, sizeof(map));                               //map[]初始化，两城镇如果不通，公路负重记为-1
        memset(name, 0, sizeof(name));
        for(int h = 0; h < m; h++)
        {
            scanf("%s%s%d", a, b, &weight);
            int t1 = find(a);
            int t2 = find(b);
            map[t1][t2] = map[t2][t1] = weight;
        }
        for(int k = 1; k <= n; k++)
            for(int j = 1; j <= n; j++)
                for(int i = 1; i <= n; i++)
                    map[i][j] = max( map[i][j], min(map[i][k], map[k][j]) );       //min(map[i][k], map[k][j])是可替代路的最大承压

        printf("Scenario #%d\n", cases++);
        scanf("%s%s", a, b);
        int t1 = find(a);
        int t2 = find(b);
        printf("%d tons\n\n", map[t1][t2]);
    }
    return 0;
}