PAT 甲级 1030 Travel Plan (30 分)(dijstra,较简单,但要注意是从0到n-1)
A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then Mlines follow, each provides the information of a highway, in the format:
City1 City2 Distance Cost
where the numbers are all integers no more than 500, and are separated by a space.
Output Specification:
For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.
Sample Input:
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
Sample Output:
0 2 3 3 40
题解:
第一行是四个数字,n,m,s,d,分别代表有n个城市,有m条路,起点城市s和终点城市d;
后面跟着m行数据,代表道路的情况,每一行四个数字,City1 City2 Distance Cost,让你求出来由起点到终点的最短的路径,如果距离相等时选择花费最小的道路,最后输出道路和最短的距离和花费;
这道题也是用dijkstra算法来进行求解,记得在最短路径的判断中加入对于花费的判断(即,当距离一样时,选择花费更小的路径);
对于路径的存储,用一个数组,每一次路径变化时,只需要存下来这个城市是由哪个城市来的,最后一个栈从终点城市回溯输出即可。
第一次没过:
没注意到城市是从0到n-1,我太粗心直接写成了1到n编号,后来才发现。
这道题没有重边的情况,但是要注意将来会不会遇到重边的特殊情况。
AC代码:
#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int n,m,s,dis;
int e[][];//距离
int d[];//距离
int dc[];//花费
int ec[][];//花费
bool v[];//标记有没有遍历过
int p[];//记录路径
void dijstra(int s){
memset(v,,sizeof(v));
for(int i=;i<=n;i++){
d[i]=e[s][i];
dc[i]=ec[s][i];
p[i]=s;
}
v[s]=;
for(int i=;i<=n;i++){
int k=-;
int mi=inf;
for(int j=;j<=n;j++){
if(mi>d[j]&&!v[j]){
k=j;
mi=d[j];
}
}
if(k==-){
break;
}
v[k]=;
for(int j=;j<=n;j++){
if(d[j]>d[k]+e[k][j]){
d[j]=d[k]+e[k][j];
dc[j]=dc[k]+ec[k][j];
p[j]=k;
}else if(d[j]==d[k]+e[k][j]){
if(dc[j]>dc[k]+ec[k][j]){
dc[j]=dc[k]+ec[k][j];
p[j]=k;
} }
}
}
}
int main()
{
cin>>n>>m>>s>>dis;
s++;//我是以1-n编号
dis++;
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
if(i==j){
e[i][j]=ec[i][j]=;
}else{
e[i][j]=ec[i][j]=inf;
}
}
}
for(int i=;i<=m;i++){
int u,v,dd,cc;
cin>>u>>v>>dd>>cc;
u++;//我是以1-n编号
v++;
e[u][v]=e[v][u]=dd;
ec[u][v]=ec[v][u]=cc;
}
dijstra(s);
//输出
stack<int>sta;
while(!sta.empty()) sta.pop();
int pair=dis;
while(pair!=s){
sta.push(pair);
pair=p[pair];
}
cout<<s-<<" ";
while(!sta.empty()){
cout<<sta.top()-<<" ";
sta.pop();
}
cout<<d[dis]<<" "<<dc[dis];
return ;
}
PAT 甲级 1030 Travel Plan (30 分)(dijstra,较简单,但要注意是从0到n-1)的更多相关文章
- PAT 甲级 1072 Gas Station (30 分)(dijstra)
1072 Gas Station (30 分) A gas station has to be built at such a location that the minimum distance ...
- 【PAT甲级】1030 Travel Plan (30 分)(SPFA,DFS)
题意: 输入N,M,S,D(N,M<=500,0<S,D<N),接下来M行输入一条边的起点,终点,通过时间和通过花费.求花费最小的最短路,输入这条路径包含起点终点,通过时间和通过花费 ...
- PAT A 1030. Travel Plan (30)【最短路径】
https://www.patest.cn/contests/pat-a-practise/1030 找最短路,如果有多条找最小消耗的,相当于找两次最短路,可以直接dfs,数据小不会超时. #incl ...
- PAT 甲级 1030 Travel Plan
https://pintia.cn/problem-sets/994805342720868352/problems/994805464397627392 A traveler's map gives ...
- PAT Advanced 1030 Travel Plan (30) [Dijkstra算法 + DFS,最短路径,边权]
题目 A traveler's map gives the distances between cities along the highways, together with the cost of ...
- 1030 Travel Plan (30分)(dijkstra 具有多种决定因素)
A traveler's map gives the distances between cities along the highways, together with the cost of ea ...
- PAT-1030 Travel Plan (30 分) 最短路最小边权 堆优化dijkstra+DFS
PAT 1030 最短路最小边权 堆优化dijkstra+DFS 1030 Travel Plan (30 分) A traveler's map gives the distances betwee ...
- [图算法] 1030. Travel Plan (30)
1030. Travel Plan (30) A traveler's map gives the distances between cities along the highways, toget ...
- PAT 甲级 1080 Graduate Admission (30 分) (简单,结构体排序模拟)
1080 Graduate Admission (30 分) It is said that in 2011, there are about 100 graduate schools ready ...
随机推荐
- 【产品对比】Word开发工具Aspose.Words和Spire.Doc性能和优劣对比一览
转:evget.com/article/2018/4/3/27885.html 概述:Microsoft Office Word是微软公司的一个文字处理器应用程序,作为办公软件必不可少的神器之一,Wo ...
- ValueError: Expecting property name: line 1 column 2 (char 1)
代码: import json str2 = '{"domain":"456"}' str1 = "{'domain':'123'}" pr ...
- Java8-Optional-No.02
import java.util.Optional; import java.util.function.Supplier; public class Optional2 { static class ...
- Java8-Lambda-No.04
public class Lambda4 { static int outerStaticNum; int outerNum; void testScopes() { int num = 1; Lam ...
- 通过德鲁伊druid给系统增加监控
系统在线上运行了一段时间后,比如一年半载的,我们发现系统可能存在某些问题,比如执行系统变慢了,比如某些spring的bean无法监控各种调用情况. 触发到db的各种执行情况,这个时候,我们就需要一个工 ...
- scrapy3 中间件的使用
前情提要: 补充知识点: ua请求头库的使用 安装: pip install fake-useragent 使用: from fake_useragent import UserAgent ua = ...
- scrapy 学习笔记2 数据持久化
前情提要:校花网爬取,并进行数据持久化 数据持久化操作 --编码流程: 1:数据解析 2:封装item 类 3: 将解析的数据存储到实例化好的item 对象中 4:提交item 5:管道接收item然 ...
- MySQL sql join 算发
在MySQL中,可以使用批量密钥访问(BKA)连接算法,该算法使用对连接表的索引访问和连接缓冲区. BKA算法支持:内连接,外连接和半连接操作,包括嵌套外连接. BKA的优点:更加高效的表扫描提高了连 ...
- DP(第二版)
第一版请见:直通 话不多说,直接上题 1.P1040 加分二叉树 直通 思路: 已知中序遍历,相当于一段区间了,所以我们枚举一个k,如果以k为根节点,能够将分数更新,那么这段区间的根节点就置为k,最后 ...
- read,write,lseek
转自 http://blog.csdn.net/todd911/article/details/11237627 1.read 调用read函数从文件去读数据,函数定义如下: #include < ...