朴素版dijkstra问题

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

const int N = 220;
int g[N][N], dist[N];
bool st[N];
int n, m;

int dijkstra()  // 求1号点到n号点的最短路距离，如果从1号点无法走到n号点则返回-1
{
    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()
{
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);
    }

    int t = dijkstra();
    cout << t << endl;
    return 0;

}

堆优化版dijkstra算法

#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] > dist[ver] + w[i])
            {
                dist[j] = dist[ver] + w[i];
                heap.push({dist[j], j});
            }
        }
    }

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

int main()
{
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);
    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }

    cout << dijkstra() << endl;

    return 0;
}

bellman_ford算法

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

const int N = 510, M = 10010;
int dist[N], backup[N];
int n, m, k;

struct Edge
{
    int a, b, w;
}edges[M];

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

    for(int i = 0; i < k; i ++ )
    {
        memcpy(backup, dist, sizeof dist);
        for(int j = 0; j < m; j ++ )
        {
            int a = edges[j].a, b = edges[j].b, w = edges[j].w;
            dist[b] = min(dist[b], backup[a] + w);
        }
    }
    if(dist[n] == 0x3f3f3f3f / 2)   return -1;
    return dist[n];
}

int main()
{
    cin >> n >> m >> k;
    for(int i = 0; i < m; i ++ )
    {
        int a, b , w;
        cin >> a >> b >> w;
        edges[i] = {a, b, w};
    }

    int t = bellman_ford();
    if(t == -1) cout << "NO" << endl;
    cout << t << endl;
    return 0;
}

spfa算法

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

using namespace std;

const int N = 220;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];
int n, m;

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

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

    queue<int> q;
    q.push(1);
    st[1] = true;

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        st[t] = false;

        for(int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                if(!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    if(dist[n] == 0x3f3f3f3f)   return -1;
    return dist[n];
}

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

    memset(h, -1, sizeof h);

    while(m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    int t = spfa();
    if(t == -1)     puts("impossible");
    else    cout << t << endl;

    return 0;
}

spfa算法判断是否存在负环

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

using namespace std;

const int N = 220;
int h[N], w[N], e[N], ne[N], idx;
int dist[N], cnt[N];
bool st[N];
int n, m, k;

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

int spfa()
{
    queue<int> q;
    for(int i = 1; i <= n; i ++ )
    {
        st[i] = true;
        q.push(i);
    }

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        st[t] = false;

        for(int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                cnt[j] = cnt[t] + 1;
                if(cnt[j] >= n) return true;
                if(!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m -- )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    if(spfa())  cout << "YES" << endl;
    else    cout << "NO" << endl;
    return 0;
}

floyd算法————多源汇最短路

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

const int N = 220, INF = 1e9;
int d[N][N];
int n, m, Q;

void floyd()
{
    for(int k = 1; k <= n; k ++ )
        for(int i = 1; i <= n; i ++ )
            for(int j = 1; j <= n; j ++ )
            d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    cin >> n >> m >> Q;
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= n; j ++ )
            if(i == j)  d[i][j] = 0;    //如果存在环
            else    d[i][j] = INF;

    while (m -- )
    {
        int a, b, w;
        cin >> a >> b >> w;
        d[a][b] = min(d[a][b], w);
    }

    floyd();

    while(Q -- )
    {
        int x, y;
        cin >> x >> y;

        if(d[x][y] > INF / 2)  puts("impossible");

        else    cout << d[x][y] << endl;
    }
    return 0;
}