「AcWing学习记录」Dijkstra

AcWing 849. Dijkstra求最短路 I

原题链接

朴素Dijkstra
1.dis[1] = 0, dis[i] = \(+\infty\)
2.for(int i = 0; i < n; i++)
s：当前已确定最短距离的点
t \(\leftarrow\) 不在s中的距离最近的点
s \(\leftarrow\) t
用t更新其他点的距离

#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

const int N = 510;

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

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(!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;

        st[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()
{
    scanf("%d%d", &n, &m);

    memset(g, 0x3f, sizeof g);

    while(m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        g[a][b] = min(g[a][b], c);
    }

    int t = dijkstra();

    printf("%d", t);

    return 0;
}

AcWing 850. Dijkstra求最短路 II

原题链接