Heavy Transportation
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 26968 | Accepted: 7232 |
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
题目大意，有n个城m条边，每个边有个最大的通过量，求1城市到n城市的一条最大通路容量是多少
迪杰斯特拉算法的变形，松弛条件改为道路容量为道路上容量最小的边，然后在选容量最大的路
ac代码如下：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<memory.h>
using namespace std;
long map[][];
long dp[],n;
bool v[];
void dij(int ii){
for(int i=;i<=n;i++){
dp[i]=map[ii][i];
}
dp[ii]=;v[ii]=;
int T=n;
while(T--){
int k=-,s;
for(int i=;i<=n;i++){//找下一条边
if(dp[i]>k&&!v[i]){
k=dp[i];
s=i;
}
}
v[s]=;
if(s==n)return;
for(int i=;i<=n;i++){//利用下一条边进行松弛
if(!v[i]&&dp[i]<min(dp[s],map[s][i])){
dp[i]=min(dp[s],map[s][i]);
}
}
}
}
int main(){
long T,m,s,e,c,ca=;
cin>>T;
while(T--){
cin>>n>>m;
memset(v,,sizeof(v));
memset(dp,,sizeof(dp));
memset(map,,sizeof(map));
for(int i=;i<=m;i++){
cin>>s>>e>>c;
map[s][e]=map[e][s]=max(map[s][e],c);
}
dij();
cout<<"Scenario #"<<ca++<<":"<<endl;
cout<<dp[n]<<endl<<endl;
}
return ;
}
提交结果：
