dijkstra

不能求带负权的最短路。

朴素版

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

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;
    else return dist[n];

}
int main()
{
    cin >> n>>m;
    memset(g,0x3f,sizeof g);
    for(int i=1;i<=m;i++)
    {
        int a,b,c;
        cin >> a>>b>>c;
        g[a][b]=min(g[a][b],c);
    }

    cout << dijkstra()<<endl;
    return 0;
}

堆优化版

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 1e6+10;
int h[N],e[N],ne[N],w[N],idx;
int 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 dijkstra()
{
    priority_queue<PII,vector<PII>,greater<PII>> heap;
    memset(dist,0x3f,sizeof dist);
    dist[1]=0;
    heap.push({0,1});

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

        int ver=t.y;
        if(st[ver]) continue;
        st[ver]=true;

        for(int i=h[ver];~i;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);
    }
    printf("%d\n",dijkstra());
    return 0;
}

spfa

不仅可以求带负权的连通块，还可以判断是否有负环

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

using namespace std;
const int N = 100010;
int h[N],ne[N],e[N],w[N],idx;
bool st[N];
int dist[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 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;i=ne[i])
        {
            int j=e[i];
            if(dist[j]>dist[t]+w[i])
            {
                dist[j]=dist[t]+w[i];
                if(!st[j])
                {
                    st[j]=true;
                    q.push(j);
                }
            }
        }
    }
    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==0x3f3f3f3f) puts("impossible");
    else cout << t<<endl;
    return 0;
}

判断负环

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 2010,M=10010;
int h[N],w[M],e[M], ne[M], idx;
int dist[N],cnt[N];
bool st[N];
int n,m;

void add(int a, int b, int c)  // 添加一条边a->b，边权为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 ++ )
    {
        st[i]=true;
        q.push(i);
    }
    while(q.size())
    {
        auto t=q.front();
        q.pop();
        st[t]=false;
        for (int i = h[t];~i; 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])
                {
                    st[j]=true;
                    q.push(j);
                }
            }
        }
    }
    return false;
}
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);
    }
    if(spfa()) puts("Yes");
    else puts("No");
    return 0;
}

floyd

多源最短路问题。

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

using namespace std;
const int N = 210,INF=1e9;
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]=0;
                else g[i][j]=INF;
            }
    while (m -- )
    {
        int a,b,w;
        cin >> a>>b>>w;
        g[a][b]=min(g[a][b],w);
    }
    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;
}