1030. Travel Plan (30)
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 Input4 5 0 3 0 1 1 20 1 3 2 30 0 3 4 10 0 2 2 20 2 3 1 20Sample Output
0 2 3 3 40
最短路就直接用dijkstra,稍微改一改，就可以变成这题的解法。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include <stack>
using namespace std;
int grond[502][502]; //路长
int val[502][502]; //价值
int n,m,st,en;
int vis[501]; //走过的路
int d[501]; //最短路长
int co[501]; //最少价值
int pre[501]; //路径
void djs(){ //dikjstra
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
for(int i = 0;i<n;i++)
{
d[i]=grond[st][i];
co[i]=val[st][i];
}
d[st]=0;
co[st]=0;
vis[st]=1;
while(1)
{
int k = -1;
for(int j = 0 ;j<n;j++)
{
if(!vis[j] && (k==-1||d[k]>d[j]))//找到当前没用过的最接近原点的点
{
k=j;
}
}
if(k==-1)
break;
vis[k]=1;
for(int j=0;j<n;j++)//对选中的点更新其他所有点
if(!vis[j] && d[j]>grond[k][j]+d[k]){
d[j]=grond[k][j]+d[k];
co[j]=val[k][j]+co[k];
pre[j]=k;//记录前一个数字
}
else if(!vis[j] && d[j]==grond[k][j]+d[k] && co[j]>val[k][j]+co[k]){
co[j]=val[k][j]+co[k];
pre[j]=k;
}
}
}
int main()
{
cin>>n>>m>>st>>en;
memset(grond,999999,sizeof(grond));
memset(val,999999,sizeof(val));
for(int i = 0 ;i<m;i++){
int a,b,c,d;
cin>>a>>b>>c>>d;
grond[a][b]=grond[b][a]=c;
val[a][b]=val[b][a]=d;
}
djs();
stack<int> ans;//对于路径倒序输出
int tem = en;
while(tem!=-1)
{
ans.push(tem);
tem = pre[tem];
}
cout<<st<<" ";
while(!ans.empty())
{
cout<<ans.top()<<" ";
ans.pop();
}
cout<<d[en]<<" "<<co[en]<<endl;
return 0;
}