hdu-3790最短路刷题

Posted 2021-01-14 31415926535x

---

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)