有代价的单源最短路径

问题：有代价的单源最短路径，并要求存储路径。（求最短的路径，并使代价最小）

特点：

* 存储路径：决定了难以用dijkstra，可以用flody，用path[i][j]表示 i 想走到 j 迈出的第一步。假设k是 i->j 的中间节点，更新时候用path[i][j] = path[i][k]，具体做法见link。但是flody比较耗时（O（N^3））

* 有代价：如果想用flody的话，有要求代价最小，就需要将最短路相等的都记录下来。边一多代价更上去了。所以还是dfs比较方便。

 

例题：

PAT 1018（http://pat.zju.edu.cn/contests/pat-a-practise/1018）

 

这道题关键要想到用dfs，就好做了。Code：

 

 

#include<iostream>
#include<memory.h>
#include<vector>
using namespace std;
#define N 505
int map[N][N];
vector<int> path,min_path;
int bike[N];
bool visit[N];
#define INF 10000000
int dest,n,req;//req:require
int min_d,min_bring,min_back;
int cur_d,cur_bring,cur_back;

void dfs(int p)
{
    int i;
    if(p==dest)
    {
        if(cur_d<min_d ||
            cur_d==min_d && cur_bring < min_bring||
            cur_d == min_d && cur_bring == min_bring && cur_back<min_back)
        {
            min_d = cur_d;
            min_bring = cur_bring;
            min_back = cur_back;
            min_path = path;
        }
        return;
    }
    else
    {
        for(i=0;i<=n;i++)
        {
            if(!visit[i]&&map[p][i]!=INF)
            {
                cur_d += map[p][i];
                visit[i] = true;
                path.push_back(i);
                if(bike[i]<req)
                {
                    if(cur_back >= req-bike[i])
                    {
                        cur_back -= (req-bike[i]);
                        dfs(i);
                        cur_back += (req-bike[i]);
                    }
                    else
                    {
                        int t = cur_back;
                        cur_bring += req-bike[i] - cur_back;
                        cur_back = 0;
                        dfs(i);
                        cur_back = t;
                        cur_bring -= req-bike[i] - cur_back;
                    }
                }
                else
                {
                    cur_back += bike[i]-req;
                    dfs(i);
                    cur_back -= (bike[i]-req);
                }
                path.pop_back();
                visit[i] = false;
                cur_d-=map[p][i];
            }
        }
    }
}

int main()
{
    int cm,m,i,j,t,a,b;
    scanf("%d%d%d%d",&cm,&n,&dest,&m);
    req = cm/2;
    for(i=0;i<n;i++)
        scanf("%d",&bike[i+1]);
    memset(visit,false,sizeof(visit));
    visit[0] = true;

    for(i=0;i<=n;i++)
        for(j=0;j<=n;j++)
            map[i][j] = INF;

    for(i=0;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&t);
        if(map[a][b]>t)
            map[a][b] = map[b][a] = t;
    }
    min_d = min_bring = min_back = INF;
    cur_d = cur_bring = cur_back = 0;
    path.push_back(0);
    dfs(0);
    cout<<min_bring<<" ";
    printf("%d",min_path[0]);
    for(i=1;i<min_path.size();i++)
        printf("->%d",min_path[i]);
    cout<<" "<<min_back<<endl;
    return 0;
}