POJ2263 Heavy Cargo

Heavy Cargo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2412		Accepted: 1321

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

思路：给出N个地点，然后给出这些地点之间的路径可以承受的汽车最大的压力。首先将这N个地点映射为数字。然后建立矩阵。最后可以用求最短路径的方法进行变形，从而求出答案。

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <map>

#define MAXINT 99999999

using namespace std;

int data[204][204];
int dis[204];
int isVis[204];

char maps[204][34];
int length=0;

int findv(char str[34])
{
int i;
for(i=1;i<length;i++)
{
if(strcmp(maps[i],str)==0)
return i;
}

strcpy(maps[length],str);
length++;

return length-1;
}

int minn(int x,int y)
{if(x>y)
return y;
return x;
}

int main(int argc, char *argv[])
{

int i,j,k;
int n,m;
int ncount=0;

while(scanf("%d%d",&n,&m)!=EOF)
{
if((n==0)&&(m==0))
break;

length=1;

getchar();

ncount++;

for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
data[i][j]=0;

for(i=1;i<=m;i++)
{
char s1[34],s2[34];
int w;

scanf("%s%s%d",s1,s2,&w);

getchar();

int v1=findv(s1);
int v2=findv(s2);

data[v1][v2]=data[v2][v1]=w;
}

char s1[34],s2[34];

scanf("%s%s",s1,s2);
getchar();

int startv=findv(s1);
int endv=findv(s2);

for(i=1;i<=n;i++)
isVis[i]=0;

for(i=1;i<=n;i++)
{dis[i]=data[startv][i];}

int maxdis=0;
isVis[startv]=1;

for(i=1;i<=n;i++)
{
k=-1;
maxdis=0;

for(j=1;j<=n;j++)
{
if((isVis[j]==0)&&(maxdis<dis[j]))
{k=j;maxdis=dis[j];}
}

isVis[k]=1;

for(j=1;j<=n;j++)
{
int tmp=minn(dis[k],data[k][j]);
if((isVis[j]==0)&&(dis[j]<tmp))
{dis[j]=tmp;}
}
}

printf("Scenario #%d\n",ncount);

printf("%d tons\n\n",dis[endv]);

}

//system("PAUSE");
return EXIT_SUCCESS;
}