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)