「AcWing学习记录」Dijkstra

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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

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

using namespace std;

typedef pair<int, int> PII;

const int N = 1e6 + 10;

int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;

    priority_queue<PII, vector<PII>, greater<PII> > heap;
    heap.push({0, 1});

    while(heap.size())
    {
        auto t = heap.top();
        heap.pop();

        int ver = t.second, distance = t.first;
        if(st[ver]) continue;
        st[ver] = true;

        for(int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(dist[j] > distance + w[i])
            {
                dist[j] = distance + w[i];
                heap.push({dist[j], j});
            }
        }
    }

    if(dist[n] == 0x3f3f3f3f) return -1;
    else return dist[n];
}

int main()
{
    cin >> n >> m;

    memset(h, -1, sizeof h);

    while(m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    int t = dijkstra();

    cout << t << endl;

    return 0;
}
posted @ 2023-02-19 20:25  恺雯  阅读(29)  评论(0编辑  收藏  举报