F - Heavy Transportation
Time Limit:3000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 1797
Description
Background
Hugo
Heavy is happy. After the breakdown of the Cargolifter project he can
now expand business. But he needs a clever man who tells him whether
there really is a way from the place his customer has build his giant
steel crane to the place where it is needed on which all streets can
carry the weight.
Fortunately he already has a plan of the city with all
streets and bridges and all the allowed weights.Unfortunately he has no
idea how to find the the maximum weight capacity in order to tell his
customer how heavy the crane may become. But you surely know.
Problem
You are given the plan of the city, described
by the streets (with weight limits) between the crossings, which are
numbered from 1 to n. Your task is to find the maximum weight that can
be transported from crossing 1 (Hugo‘s place) to crossing n (the
customer‘s place). You may assume that there is at least one path. All
streets can be travelled in both directions.
Input
The first line contains the number of scenarios (city plans). For each
city the number n of street crossings (1 <= n <= 1000) and number m
of streets are given on the first line. The following m lines contain
triples of integers specifying start and end crossing of the street and
the maximum allowed weight, which is positive and not larger than
1000000. There will be at most one street between each pair of
crossings.
Output
The output for every scenario begins with a line containing "Scenario
#i:", where i is the number of the scenario starting at 1. Then print a
single line containing the maximum allowed weight that Hugo can
transport to the customer. Terminate the output for the scenario with a
blank line.
Sample Input
1 3 3 1 2 3 1 3 4 2 3 5
Sample Output
Scenario #1: 4 每条路都有一个限制的重量 求从1到n最多可以装载多少货物顺利通过
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
//const int inf=0x7fffffff;
const int MAXN=1010;
//#define typec int
const int INF=0x3f3f3f3f;//防止后面溢出，这个不能太大
bool vis[MAXN];
int dis[MAXN];
int map[MAXN][MAXN];
int n;
void Dijkstra(int beg)
{
for(int i=1; i<=n; i++)
{
dis[i]=map[beg][i];
vis[i]=false;
}
dis[beg]=0;
for(int j=0; j<n; j++)
{
int k=-1;
int Min=-1;
for(int i=1; i<=n; i++)
if(!vis[i]&&dis[i]>Min)
{
Min=dis[i];
k=i;
}
if(k==-1)
break;
vis[k]= true;
for(int i=1; i<=n; i++)
if(!vis[i]&&dis[i]<min(dis[k],map[i][k]))
{
dis[i]=min(dis[k],map[i][k]);
}
}
}
int main(){
int t;
scanf("%d",&t);
int cnt=0;
while(t--){
cnt++;
int m;
memset(vis,false,sizeof(vis));
scanf("%d%d",&n,&m);
/*for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
if(i==j)
map[i][i]=0;
else
map[i][j]=map[j][i]=INF;
}
}*/
memset(map,0,sizeof(map));
int u,v,w;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&u,&v,&w);
map[u][v]=map[v][u]=w;
}
Dijkstra(1);
printf("Scenario #%d:\n",cnt);
printf("%d\n",dis[n]);
puts("");
}
return 0;
}