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  • 题目1008:最短路径问题(最短路径问题dijkstra算法)

    题目链接:http://ac.jobdu.com/problem.php?pid=1008

    详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus

    参考代码:

    //
    //  1008 最短路径问题.cpp
    //  Jobdu
    //
    //  Created by PengFei_Zheng on 19/04/2017.
    //  Copyright © 2017 PengFei_Zheng. All rights reserved.
    //
     
    #include <stdio.h>
    #include <iostream>
    #include <algorithm>
    #include <string.h>
    #include <cmath>
    #include <vector>
     
    #define MAX_SIZE 1010
     
    using namespace std;
     
    int n, m;
     
     
    struct E{//the adjective matrix
        int next;
        int len;
        int cost;
    };
     
    vector<E> edge[MAX_SIZE];
     
    int dis[MAX_SIZE];//dis[i] keep the length from start to [i] so and cost[i],mark[i]
    int cost[MAX_SIZE];
    bool mark[MAX_SIZE];
     
    int main(){
        while(scanf("%d%d",&n,&m)!=EOF){
            if(0==n && 0==m) break;
            for(int i = 1 ; i <= n ; i++){
                edge[i].clear();//clear the matrix
            }
            for(int i = 1 ; i <= m ; i++){
                int a, b, len, cost;
                scanf("%d%d%d%d",&a,&b,&len,&cost);
                E tmp;
                tmp.len = len;
                tmp.cost = cost;
                tmp.next = b;
                edge[a].push_back(tmp);//the node a has those info and the next node is b
                tmp.next = a;
                edge[b].push_back(tmp);//the node b has those info and the next node is a
            }
             
            int s, t;
            scanf("%d%d",&s,&t);//start node and end node
             
            for(int i = 1 ; i <= n ; i++){//init
                dis[i] = -1;
                mark[i] = false;
                cost[i] = 0;
            }
             
            dis[s]=0;
            mark[s] = true;//whether this node is in the aggregate
            int newP = s;//new dot needed to add
            for(int i = 1 ; i < n ; i++){
                for(int j = 0 ; j <edge[newP].size(); j++){//all the nodes in the newP
                    //get info
                    int next = edge[newP][j].next;
                    int len = edge[newP][j].len;
                    int cos = edge[newP][j].cost;
                     
                    if(mark[next] == true) continue;//already in the aggreagte
                    //can't reach or has a less length or has smaller cost
                    if(dis[next]==-1 || dis[next] > dis[newP]+len || (dis[next]==dis[newP]+len && cost[next] > cost[newP]+cos)){
                        dis[next] = dis[newP] + len;
                        cost[next] = cost[newP] + cos;
                    }
                }
                int min = 2121212121;
                for(int j = 1 ; j <= n ; j++){//find a new node to add
                    if(mark[j]==true) continue;//not in the aggregate
                    if(dis[j] == -1) continue;//can't  reach
                    if(dis[j]<min){
                        min = dis[j];
                        newP = j;
                    }
                }
                mark[newP] = true;
            }
            printf("%d %d
    ",dis[t],cost[t]);
        }
        return 0;
    }
    /**************************************************************
        Problem: 1008
        User: zpfbuaa
        Language: C++
        Result: Accepted
        Time:10 ms
        Memory:1552 kb
    ****************************************************************/
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  • 原文地址:https://www.cnblogs.com/zpfbuaa/p/6736356.html
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