PAT甲级1111 Online Map (30分)|C++实现

一、题目描述

原题链接
Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.

Input Specification:

​​Output Specification:

Sample Input 1:

10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
3 4 0 3 2
3 9 1 4 1
0 6 0 1 1
7 5 1 2 1
8 5 1 2 1
2 3 0 2 2
2 1 1 1 1
1 3 0 3 1
1 4 0 1 1
9 7 1 3 1
5 1 0 5 2
6 5 1 1 2
3 5

Sample Output 1:

Distance = 6: 3 -> 4 -> 8 -> 5
Time = 3: 3 -> 1 -> 5

Sample Input 2:

7 9
0 4 1 1 1
1 6 1 1 3
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 1 3
3 2 1 1 2
4 5 0 2 2
6 5 1 1 2
3 5

Sample Output 2:

Distance = 3; Time = 4: 3 -> 2 -> 5

二、解题思路

这道题不难但是有点复杂，利用了两次DIjkstra算法，还有就是利用pre数组和dfs还原路径。代码很长但是比较易懂，具体细节可参考注释。

三、AC代码

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int INF = 999999999;
const int maxn = 510;
int dis[maxn], Time[maxn], e[maxn][maxn], w[maxn][maxn], dispre[maxn], Timepre[maxn], weight[maxn], nodeNum[maxn];
bool vis[maxn];
vector<int> dispath, Timepath, temppath;
int st, fin, minnode = INF;
void dfsdispath(int v)  //存放距离最短的路径
{
    dispath.push_back(v);
    if(v == st) return;
    dfsdispath(dispre[v]);
}
void dfsTimepath(int v) //存放时间最短的路径
{
    Timepath.push_back(v);
    if(v == st) return;
    dfsTimepath(Timepre[v]);
}
int main()
{
    fill(dis, dis+maxn, INF);
    fill(Time, Time+maxn, INF);
    fill(weight, weight+maxn, INF);
    fill(e[0], e[0]+maxn*maxn, INF);
    fill(w[0], w[0]+maxn*maxn, INF);
    int n, m;
    scanf("%d%d", &n, &m);
    int a, b, flag, len, t;
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%d%d%d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1)
        {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d%d", &st, &fin);   //输入起点终点
    dis[st] = 0;    //初始化起点的距离
    for(int i=0; i<n; i++)  dispre[i] = i;  //初始化dispre数组，每个结点的前一个结点设为自身
    for(int i=0; i<n; i++)  //开始Dijkstra算法
    {
        int u = -1, minn = INF;
        for(int j=0; j<n; j++)  //找到没有访问过的距离最短的结点
        {
            if(!vis[j] && dis[j] < minn)
            {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        vis[u] = true;
        for(int v=0; v<n; v++)
        {
            if(!vis[v] && e[u][v] != INF)
            {
                if(e[u][v] + dis[u] < dis[v])   //更新距离
                {
                    dis[v] = e[u][v] + dis[u];
                    dispre[v] = u;
                    weight[v] = weight[u] + w[u][v];
                }
                else if(e[u][v] + dis[u] == dis[v] && weight[u] + w[u][v] < weight[v])  //距离相等但是时间更短，更新
                {
                    weight[v] = weight[u] + w[u][v];
                    dispre[v] = u;
                }                
            }
        }
    }
    dfsdispath(fin);
    Time[st] = 0;
    fill(vis, vis+maxn, false);
    for(int i=0; i<n; i++)  //Dijkstra算法开始
    {
        int u = -1, minn = INF;
        for(int j=0; j<n; j++)  //找到时间最短的点
        {
            if(!vis[j] && Time[j] < minn)
            {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        vis[u] = true;
        for(int v=0; v<n; v++)
        {
            if(!vis[v] && w[u][v] != INF)
            {
                if(w[u][v] + Time[u] < Time[v])
                {
                    Time[v] = Time[u] + w[u][v];
                    Timepre[v] = u;
                    nodeNum[v] = nodeNum[u] + 1;
                }
                else if(Time[u] + w[u][v] == Time[v] && nodeNum[u]+1 < nodeNum[v])
                {
                    Timepre[v] = u;
                    nodeNum[v] = nodeNum[u] + 1;
                }
            }
        }
    }
    dfsTimepath(fin);   //存放时间最短路径
    printf("Distance = %d", dis[fin]);
    if(dispath == Timepath) printf("; Time = %d: ", Time[fin]); //可以直接判断两vector是否相等
    else
    {
        printf(": ");
        for(int i=dispath.size()-1; i>=0; i--)  //从后往前输出
        {
            printf("%d", dispath[i]);
            if(i!=0)    printf(" -> ");
        }
        printf("\nTime = %d: ", Time[fin]);
    }
    for(int i=Timepath.size() - 1; i>=0; i--)
    {
        printf("%d", Timepath[i]);
        if(i!=0)    printf(" -> ");
    }
    return 0;
}