雖然floyid算法效率不高，三重循環，但是比較容易理解，通過適當的剪支還是可以 提高效率的。但是比賽時間有限，而且時間給的也夠，因此我的算法的時間複雜度是O(n*n*n)。
我的代碼：
/*
Heavy Cargo
Time Limit:1000MS Memory Limit:65536K
Total Submit:225 Accepted:140
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
Source
Ulm Local 1998
*/
#include "iostream"
#include "memory"
using namespace std;
int main()
{
int n,r,w,dis[200][200],t=1,flag=0,i,j,k,z=0,min,max;
char ch[30];
char p[200][30];
while(cin>>n>>r&&n>=2&&n<=200&&r>=1&&r<=19900)
{
memset(p,'a',sizeof(p));
memset(dis,0,sizeof(dis));
for(;r-1>=0;r--)
{
cin>>ch;
if(p[0][0]=='a')
{
strcpy(p[0],ch);
z++;
}
for(i=0;i<z;i++)
{
if(!strcmp(p[i],ch))
{
flag=1;
break;
}
}
if(!flag)
{
i++;
strcpy(p[z],ch);
z++;
}
flag=0;
cin>>ch;
for(j=0;j<z;j++)
{
if(!strcmp(p[j],ch))
{
flag=1;
break;
}
}
if(!flag)
{
strcpy(p[z],ch);
z++;
}
flag=0;
cin>>w;
dis[i][j]=w;
dis[j][i]=dis[i][j];
}
for(k=0;k<n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
if(dis[i][k]>dis[k][j])
min=dis[k][j];
else
min=dis[i][k];
if(dis[i][j]>min)
max=dis[i][j];
else
max=min;
dis[i][j]=max;
}
cin>>ch;
for(i=0;i<z;i++)
{
if(!strcmp(p[i],ch))
{
break;
}
}
cin>>ch;
for(j=0;j<z;j++)
{
if(!strcmp(p[j],ch))
{
break;
}
}
cout<<"Scenario #"<<t<<endl;
cout<<dis[i][j]<<" tons"<<endl;
t++;
z=0;
}
return 0;
}