最短路问题总结

Dijkstra-朴素O(n^2)

1. 初始化距离数组, dist[1] = 0, dist[i] = inf;
2. for n次循环 每次循环确定一个min加入S集合中，n次之后就得出所有的最短距离
3. 不在S中dist_min的点->t
4. t->S加入最短路集合
5. 用t更新到其他点的距离

Dijkstra-堆优化O(mlogm)

1. 利用邻接表，优先队列
2. 在priority_queue,greater> heap;中将返回堆顶
3. 利用堆顶来更新其他点，并加入堆中类似宽搜

Bellman_ford O(nm)

1. 注意连锁想象需要备份, struct Edge{int a,b,c} Edge[M];
2. 初始化dist, 松弛 dist[x.b] = min(dist[x.b], backup[x.a]+x.w);
3. 松弛k次，每次访问m条边

Spfa O(n)～O(nm)

1. 利用队列优化仅加入修改过的地方
2. for k次
3. for 所有边利用宽搜模型去优化bellman_ford算法
4. 更新队列中当前点的所有出边

Floyd O(n3)

1. 初始化d
2. k, i, j 去更新d

Dijstra朴素版：

#include 
#include 
using namespace std;
const int N = 510;
int g[N][N], dist[N];
bool st[N];
int n, m;

int Dijstra()
{
    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;
        }
        
        for(int j = 1; j <= n; j++)
        {
            if(dist[j] > dist[t] + g[t][j])
            {
                dist[j] = dist[t] + g[t][j];
            }
        }
        
        st[t] = true;
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    else 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);
    }
    
    cout << Dijstra() << endl;
    
    return 0;
}

Dijstra堆优化版：

#include 
#include 
#include 
using namespace std;
const int N = 200010;
typedef pair<int,int>PII; 
int e[N], ne[N], w[N], h[N], idx, dist[N];
bool st[N];
int n, m;

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

int Dijstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    priority_queue, greater > heap;
    heap.push({0,1});
    
    while(heap.size())
    {
        PII 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;
    else 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);
    }
    
    cout << Dijstra() << endl;
    
    return 0;
}

bellman_Ford：

#include 
#include 
using namespace std;
const int N = 100010;
struct Edge
{
    int a, b, c;
}edge[N];
int dist[N], backup[N];
int n, m, k;

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 = 1; j <= m; j++)
        {
            dist[edge[j].b] = min(dist[edge[j].b], backup[edge[j].a] + edge[j].c);
        }
    }
}

int main()
{
    cin >> n >> m >> k;
    for(int i = 1; i <= m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        edge[i] = {a, b, c};
    }
    
    bellman_Ford();
    if(dist[n] > 0x3f3f3f3f / 2) puts("impossible");
    else cout << dist[n] << endl;
    
    return 0;
}

 spfa求最短路：

#include 
#include 
#include 
using namespace std;
const int N = 100010;
int e[N], ne[N], h[N], w[N], dist[N], idx;
bool st[N];
int n, m;

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, 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())
    {
        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];
                if(!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
}
int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while(m--)
    {
        int x, y, z;
        cin >> x >> y >> z;
        add(x, y, z);
    }
    
    spfa();
    
    if(dist[n] == 0x3f3f3f3f) puts("impossible");
    else cout << dist[n] << endl;
    
    return 0;
}

spfa判断负环：

#include 
#include 
#include 
using namespace std;
const int N = 100010;
int e[N], ne[N], h[N], w[N], idx, dist[N], cnt[N];
int n, m;
bool st[N];

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

bool spfa()
{
    memset(dist, 0x3f, sizeof dist);
    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;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    
    if(spfa()) puts("Yes");
    else puts("No");
    
    return 0;
}

Folyd算法：(求多源最短路）

#include 
#include 
using namespace std;
const int N = 210, INF = 0x3f3f3f3f;
int g[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++)
                g[i][j] = min(g[i][j], g[i][k] + g[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) g[i][j] = INF;
            else g[i][j] = 0;
            
    while(m--)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);
    }
    
    Floyd();
    
    while(q--)
    {
        int x, y;
        cin >> x >> y;
        if(g[x][y] > INF / 2) puts("impossible");
        else cout << g[x][y] << endl;
    }
    
    return 0;
}