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 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 (≤500) 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 M lines 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
题目大意：求起点到终点的最短路径最短距离和花费，要求首先路径最短，其次花费最少，要输出完整路径
分析：Dijksta + DFS。 Dijkstra记录路径pre数组，然后用dfs求最短的一条mincost以及它的路径path，最后输出path数组和mincost
注意路径path因为是从末端一直压入push_back到path里面的，所以要输出路径的时候倒着输出
关键是DFS的写法,应该是在一个链状的图数据结构上进行搜索,所以先加一个判断条件,如果到了起点就统计总长度,如果更小,更新anscost并更新path.注意这个时候仍然要pop_back(),没准还有一条更短的路径能够通向起点,之后dfs之后pop_back是常规操作.
void dfs(int v)
{
temppath.push_back(v);
if(v==from)
{
int tempcost=0;
for(int i=temppath.size()-1;i>=1;i--)
{
int id=temppath[i];
int nextid=temppath[i-1];
tempcost+=length[id][nextid];
}
if(tempcost<anscost)
{
path=temppath;
anscost=tempcost;
}
temppath.pop_back();//还得继续迭代呢,没准费用更小
return ;
}
for(int i=0;i<pre[v].size();i++)
{
dfs(pre[v][i]);
}
temppath.pop_back();
}
#include <iostream>
#include<bits/stdc++.h>
#define each(a,b,c) for(int a=b;a<=c;a++)
#define de(x) cout<<#x<<" "<<(x)<<endl
using namespace std;
const int maxn=500+5;
const int inf=0x3f3f3f3f;
int dis[maxn];
bool vis[maxn];
vector<int>pre[maxn];
vector<int>path,temppath;
int length[maxn][maxn];
struct Edge
{
int v;
int len;
int cost;
Edge(int v,int len,int cost):v(v),len(len),cost(cost){}
};
vector<Edge>G[maxn];
struct node
{
int v;
int len;
node(int v=0,int len=0):v(v),len(len){}
bool operator<(const node&r)const
{
return len>r.len;
}
};
void dijkstra(int n,int start)
{
each(i,0,n-1)
{
vis[i]=false;
dis[i]=inf;
}
dis[start]=0;
priority_queue<node>Q;
node temp;
Q.push(node(start,0));
while(!Q.empty())
{
temp=Q.top();
Q.pop();
int u=temp.v;
if(vis[u])continue;
vis[u]=true;
for(int i=0;i<(int)G[u].size();i++)
{
int v=G[u][i].v;
int len=G[u][i].len;
int cost=G[u][i].cost;
if(!vis[v]&&dis[v]>dis[u]+len)
{
dis[v]=dis[u]+len;
pre[v].clear();
pre[v].push_back(u);
Q.push(node(v,dis[v]));
}
else if(!vis[v]&&dis[v]==dis[u]+len)///我佛拉
{
pre[v].push_back(u);
///需不需要push呢????????
}
}
}
}
/*
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
*/
int n,m,from,to;
int anscost;
void dfs(int v)
{
temppath.push_back(v);
if(v==from)
{
int tempcost=0;
for(int i=temppath.size()-1;i>=1;i--)
{
int id=temppath[i];
int nextid=temppath[i-1];
tempcost+=length[id][nextid];
}
if(tempcost<anscost)
{
path=temppath;
anscost=tempcost;
}
temppath.pop_back();//还得继续迭代呢,没准费用更小
return ;
}
for(int i=0;i<pre[v].size();i++)
{
dfs(pre[v][i]);
}
temppath.pop_back();
}
int main()
{
anscost=inf;
cin>>n>>m>>from>>to;
while(m--)
{
int a,b,l,cost;
scanf("%d%d%d%d",&a,&b,&l,&cost);
G[a].push_back(Edge(b,l,cost));
G[b].push_back(Edge(a,l,cost));
length[a][b]=length[b][a]=cost;
}
dijkstra(n,from);
dfs(to);
for(int i=path.size()-1;i>=0;i--)
{
printf("%d ",path[i]);
}
printf("%d %d
",dis[to],anscost);
return 0;
}