POJ 1797 Heavy Transportation(最小生成树或最短路)
2016-05-24 21:41
435 查看
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
题目的意思是让你求从1到n的最短路的最小边的最大值,所以Kruskal就可以水过了,当然也可以用Dijkstra过,其实很简单。有兴趣的可以拿最短路做下,在这里先贴上一个代码,,回头会把最短路的代码补上。
下面是Kruskal的代码:
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
题目的意思是让你求从1到n的最短路的最小边的最大值,所以Kruskal就可以水过了,当然也可以用Dijkstra过,其实很简单。有兴趣的可以拿最短路做下,在这里先贴上一个代码,,回头会把最短路的代码补上。
下面是Kruskal的代码:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; struct node { int x,y,cost; }a[1000005]; int pre[1005]; int fin(int x) { if(x==pre[x]) { return x; } else { return pre[x]=fin(pre[x]); } } void join(int x,int y) { int t1=fin(x); int t2=fin(y); if(t1!=t2) { pre[t1]=t2; } } bool cmp(node p,node q) { return p.cost>q.cost; } int main() { int t,n,m,iCase=0; cin>>t; while(t--) { iCase++; cin>>n>>m; for(int i=1;i<=n;i++) { pre[i]=i; } for(int i=0;i<m;i++) { cin>>a[i].x>>a[i].y>>a[i].cost; } sort(a,a+m,cmp); int sum=0; for(int i=0;i<m;i++) { if(fin(1)==fin(n)) { break; } if(fin(a[i].x)!=fin(a[i].y)) { join(a[i].x,a[i].y); sum=a[i].cost; } } cout<<"Scenario #"<<iCase<<":"<<endl; cout<<sum<<endl<<endl; } return 0; }
相关文章推荐
- 初学ACM - 组合数学基础题目PKU 1833
- POJ ACM 1001
- POJ ACM 1002
- 1611:The Suspects
- POJ1089 区间合并
- POJ 2159 Ancient Cipher
- POJ 2635 The Embarrassed Cryptographe
- POJ 3292 Semi-prime H-numbers
- POJ 2773 HAPPY 2006
- POJ 3090 Visible Lattice Points
- POJ-2409-Let it Bead&&NYOJ-280-LK的项链
- POJ-1695-Magazine Delivery-dp
- POJ1523 SPF dfs
- POJ-1001 求高精度幂-大数乘法系列
- POJ-1003 Hangover
- POJ-1004 Financial Management
- POJ1050 最大子矩阵和
- 用单调栈解决最大连续矩形面积问题
- 2632 Crashing Robots的解决方法
- 1573 Robot Motion (简单题)