G - The Tourist Guide UVA - 10099

1.题目
G - The Tourist Guide UVA - 10099
Mr. G. works as a tourist guide. His current assignment is to take some tourists from one city to
another. Some two-way roads connect the cities. For each pair of neighboring cities there is a bus
service that runs only between those two cities and uses the road that directly connects them. Each
bus service has a limit on the maximum number of passengers it can carry. Mr. G. has a map showing
the cities and the roads connecting them. He also has the information regarding each bus service. He
understands that it may not always be possible for him to take all the tourists to the destination city
in a single trip. For example, consider the following road map of 7 cities. The edges connecting the
cities represent the roads and the number written on each edge indicates the passenger limit of the bus
service that runs on that road.
Now, if he wants to take 99 tourists from city 1 to city 7, he will require at least 5 trips and the
route he should take is : 1 - 2 - 4 - 7.
But, Mr. G. finds it difficult to find the best route all by himself so that he may be able to take all
the tourists to the destination city in minimum number of trips. So, he seeks your help.
Input
The input will contain one or more test cases. The first line of each test case will contain two integers:
N (N ≤ 100) and R representing respectively the number of cities and the number of road segments.
Then R lines will follow each containing three integers: C1, C2 and P. C1 and C2 are the city numbers
and P (P > 1) is the limit on the maximum number of passengers to be carried by the bus service
between the two cities. City numbers are positive integers ranging from 1 to N. The (R + 1)-th line
will contain three integers: S, D and T representing respectively the starting city, the destination city
and the number of tourists to be guided.
The input will end with two zeroes for N and R.
Output
For each test case in the input first output the scenario number. Then output the minimum number
of trips required for this case on a separate line. Print a blank line after the output of each test case.
Sample Input
7 10
1 2 30
1 3 15
1 4 10
2 4 25
2 5 60
3 4 40
3 6 20
4 7 35
5 7 20
6 7 30
1 7 99
0 0
Sample Output
Scenario #1
Minimum Number of Trips = 5

2.题目大意
导游需要带T个人从S到D，其中先给出各个联通点之间能够带多少个人，然后问至少需要几趟才能从S到D。

3.解题思路
我是用floyd变形，找出从S到D的所有可行道路中，能够能载人最多的一条道路，然后把这条道路中载人最少的弄出来，然后算出一共需要多少趟，其中导游是需要来回走接人的，要把导游算上。

4.floyd变形

  基本的floyd在：https://blog.csdn.net/weixin_45674799/article/details/104410012
 这道题需要改变的就是：
  if (min(graph[i][k], graph[k][j]) > graph[i][j])
                    graph[i][j] = min(graph[i][k], graph[k][j]);
  在这里找到最小的那个载人量。

5.AC代码

#include 
#include
#include
#include
#include
using namespace std;
#define maxn 150
#define maxdis 2500000
int graph[maxn][maxn];
int visit[maxn];
int n;
int a[maxn],b[maxn];
double sum[maxn];

int  flyod(int l, int r)
{
    for(int k = 1; k<=n; k++)//中间点
    {
        for(int i = 1; i<=n; i++) //起点
        {
            for(int j = 1; j<=n; j++)//终点
            {
                if (graph[i][k] == -1 || graph[k][j] == -1)
                    continue;
                if (min(graph[i][k], graph[k][j]) > graph[i][j])
                    graph[i][j] = min(graph[i][k], graph[k][j]);
            }
        }
    }
    return graph[l][r];
}
int main()
{
    int N;
    int casenum = 1;
    while(scanf("%d%d",&n,&N) == 2 && !(!n&&!N))
    {
        memset(graph,-1,sizeof(graph));

        int   cost, a, b;
        for(int i = 1; i<=N; i++)
        {
            cin>>a>>b>>cost;
            graph[a][b] = graph[b][a] = cost;
        }
        int S, D, T;
        cin>>S>>D>>T;
        printf("Scenario #%d\nMinimum Number of Trips = ",casenum++);
        if( S == D)
        {
            printf("0\n");
            continue;
        }
        int ans = flyod(S,D);
        if(T%(ans-1) == 0)
        {
            printf("%d\n",T/(ans-1));
        }
        else
        {
            printf("%d\n",T/(ans-1)+1);
        }
        printf("\n");
    }
    return 0;
}