2024-9-11

最短路问题

单元最短路问题(只有一个起点)

所有边权都是正数

朴素版Dijkstra算法 O(n^2)
堆优化版的Dijkstra算法 O(mlogn)

存在负边权

Bellman-Ford算法 O(nm)
SPFA算法 O(m)

多源汇最短路问题

Floyd算法 O(n^3)

稠密图 m == n^2 ----> 邻接矩阵
稀疏图 m == n ----> 邻接表

朴素版Dijkstra算法模板

#include <iostream>
#include <cstdio>
#include <cstring>

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()
{
    cin >> 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();

    cout << t << endl;

    return 0;
}

堆优化版的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()
{
    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);
    }

    cout << dijkstra() << endl;

    return 0;
}

Bellman-Ford算法

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int N = 510, M = 10010;

int n, m, k;
int dist[N], backup[N];

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

void 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 ++ )
        {
            auto e = edges[j];
            dist[e.b] = min(dist[e.b], backup[e.a] + e.w);
        }
    }
}

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

    for (int i = 0; i < m; i ++ )
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }

    bellman_ford();

    if (dist[n] > 0x3f3f3f3f / 2) puts("impossible");
    else cout << dist[n] << endl;

    return 0;
}

SFPA求最短路算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
const int N = 1e5 + 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 ++;
}

void 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;
                }
            }
        }
    }
}

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);
    }

    spfa();

    if (dist[n] == 0x3f3f3f3f) puts("impossible");
    else cout << dist[n] << endl;

    return 0;
}

SFPA求负边权算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
const int N = 2010, M = 10010;

int n, m;
int h[N], w[M], e[M], ne[M], idx;
int dist[N], cnt[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 ++;
}

bool spfa()
{
    queue<int> q;

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

    while (q.size())
    {
        int 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()
{
    memset(h, -1, sizeof h);

    cin >> n >> m;

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

    if (spfa()) puts("Yes");
    else puts("No");

    return 0;
}

Floyd求最短路算法

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int N = 210;

int n, m, Q;
int d[N][N];

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;
            else d[i][j] = 0x3f3f3f3f;

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

    floyd();

    while (Q --)
    {
        int x, y;
        scanf("%d%d", &x, &y);

        int t = d[x][y];
        if (t > 0x3f3f3f3f / 2) puts("impossible");
        else printf("%d\n", t);
    }

    return 0;
}