Big Truck（Dijkstra）

题目链接
带权最短路，Dijkstra模板题
一开始傻傻的用Floyd求，怎么也算不对，后来又用DFS，最后一个样例超时…最后用Dijkstra求的…
后来上网查的资料才知道Dijkstra是经典的求取带权最短路算法.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define ll long long
using namespace std;
const int N = 1e5+5;
int n,t[105],m,a,b,d;
int vis[105];
int g[105][105];
int val[105][105];
int cost[105];

void dijkstra(int u)
{
    int dist[105];
    for(int i=1; i<=n; i++)
    {
        dist[i] = g[u][i];
        if(g[u][i]!=N)
            cost[i] = t[u]+t[i];
    }
    vis[u] = 1;
    int k,minn;
    while(true)
    {
        k=-1,minn = N*5;
        for(int i=1; i<=n; i++)
        {
            if(dist[i]<minn&&!vis[i])
            {
                k=i;
                minn = dist[i];
            }
        }
        if(k==-1)
            break;
        vis[k] = 1;
        for(int i=1; i<=n; i++)
        {
            if(!vis[i])
            {
                if(dist[i]>dist[k]+g[k][i])
                {
                    dist[i] = dist[k]+g[k][i];
                    cost[i] = cost[k]+t[i];
                }
                else if(dist[i]==dist[k]+g[k][i])
                {
                    if(cost[i]<cost[k]+t[i])
                    {
                        cost[i] = cost[k]+t[i];
                    }
                }
            }
        }
    }
    if(dist[n]==N)
        cout<<"impossible"<<endl;
    else
        cout<<dist[n]<<' '<<cost[n]<<endl;
}

int main()
{
    cin>>n;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
            g[i][j] = N;
    }
    for(int i=1; i<=n; i++)
        cin>>t[i];
    cin>>m;
    for(int i=1; i<=m; i++)
    {
        cin>>a>>b>>d;
        g[a][b] = g[b][a] = d;
    }
    dijkstra(1);
    return 0;
}

标签：dist,int,Big,vis,Dijkstra,cost,Truck,include,105
来源： https://blog.csdn.net/sabr_/article/details/112133778