朴素Dijkstra

朴素Dijkstra学习笔记

适用范围

• 最短路径问题->单源最短路->所有边权都是正数
• 稠密图，即边数m与$n^2$相近
• 用邻接矩阵储存

  时间复杂度

$$O(N^2)$$
N为点数

思想解释

• dist表示点离起点的距离，s存储每个点离起点的最短值
• 1.dist[1] = 0; // 表示1到起点距离为0
• 2.遍历所有点，找到不属于s的点中离起点最近的点，记录为t
• 将其存入s中
• 用t这一点更新其他点离起点的距离

例题

https://www.acwing.com/problem/content/851/

AC代码

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 510;

int dist[N];
bool s[N];
int g[N][N];
int n, m;

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for (int i = 0; i < n; i ++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if(!s[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        s[t] = true;
        
        for (int j = 1; j <= n; j ++)
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}
int main()
{
    memset(g, 0x3f, sizeof g);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++) 
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = min(g[a][b], c);
    }
    printf("%d",dijkstra());
    
    return 0;
}