• hdu-3790最短路刷题



    title: hdu-3790最短路刷题
    date: 2018-10-20 14:50:31
    tags:

    • acm
    • 刷题
      categories:
    • ACM-最短路

    概述

    一道最短路的水题,,,尽量不看以前的代码打出来,,,熟悉一下dijkstra的格式和链式前向星的写法,,,,

    虽然是水题,,,但是一开始没考虑取费用最短的wa了一发,,,,QAQ

    分析

    链式前向星存图,,再加一个数组保存源点到每个点的费用cst[maxm],,,注意取最少的费用

    代码

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cstdlib>
    #include <queue>
    
    using namespace std;
    
    const int maxn = 1e3 + 10;
    const int maxm = 1e5 + 10;
    const int inf = 0x3f3f3f3f;
    
    int head[maxm << 1];
    bool vis[maxn];
    int dis[maxm];
    int cst[maxm];
    int cnt;
    int n , m;
    
    struct edge
    {
        int to;
        int w;
        int c;
        int last;
    }edge[maxm << 1];
    
    void addedge(int u , int v , int w , int c)
    {
        edge[cnt].to = v;
        edge[cnt].w = w;
        edge[cnt].c = c;
        edge[cnt].last = head[u];
        head[u] = cnt++;
    }
    struct node
    {
        int u;
        int w;
        node(int _u , int _w):u(_u) , w(_w){}
    
        bool operator < (const node &res) const
        {
            return w > res.w;
        }
    };
    
    void dijkstra(int n , int s)
    {
        for(int i = 1; i <= n; ++i)
            dis[i] = (i == s) ? 0 : inf;
        memset(cst , inf , sizeof cst);cst[s] = 0;
        memset(vis , false , sizeof vis);
    
        priority_queue<node> q;
    
        while(!q.empty())   q.pop();
    
        q.push(node(s , 0));
    
        while(!q.empty())
        {
            node x = q.top();q.pop();
            int u = x.u;
    
            if(vis[u])  continue;
            vis[u] = true;
    
            for(int i = head[u] ; ~i; i = edge[i].last)
            {
                int to = edge[i].to;
                int w = edge[i].w;
                int c = edge[i].c;
                if(!vis[to] && dis[u] + w <= dis[to])
                {
                    dis[to] = dis[u] + w;
                    //if(cst[u] + c < cst[to])
                        cst[to] = cst[u] + c;
                    q.push(node(to , dis[to]));
                }
            }
        }
    }
    int main()
    {
        while(scanf("%d%d" , &n , &m) && n && m)
        {
            cnt = 0;
            memset(head , -1 , sizeof head);
            int u , v , w , c;
            for(int i = 1; i <= m; ++i)
            {
                scanf("%d%d%d%d" , &u , &v , &w , &c);
                addedge(u , v , w , c);
                addedge(v , u , w , c);
            }
            int s , t;
            scanf("%d%d" , &s , &t);
    
            dijkstra(n , s);
    
            printf("%d %d
    " , dis[t] , cst[t]);
    
        }
    }
    
    //最短路相等时注意取费用最短的
    //
    //5 7
    //1 2 5 5
    //2 3 4 5
    //1 3 4 6
    //3 4 2 2
    //3 5 4 7
    //4 5 2 4
    //1 3 4 4
    //1 5
    //8 10
    
    

    差不多记住了的dijkatra的代码,,,继续继续

    (end)

    剑之所指,心之所向,身之所往!!!
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  • 原文地址:https://www.cnblogs.com/31415926535x/p/9821730.html
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