【图论】单源最短路的三种算法

题目

例题可以用洛谷P3371和P4779

思路

1.dijstra
完美做出,但条件是不能有负权边,堆优化以后复杂度是O(mlogn)。
2.bellman-ford
主要用于有负权边的情况,理论复杂度是O(nm),但队列优化以后往往远小于这个复杂度。
3.floyd
多源最短路算法,这里也拿过来一块学了,复杂度是 O(n3) O ( n 3 ) ,所以不应该用多元最短路算法floyd去求高效率的单源最短路。

代码

dijkstra-heap

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;

const int maxn = 100000+10;
const int maxm = 500000+10;

struct edge{
    int u,v,w;
    edge(int u, int v, int w):u(u), v(v), w(w){};
};

struct HeapNode{
    int d, u;
    bool operator <(const struct HeapNode& rhs) const{
        return d > rhs.d;
    }
};

int n,m,start, d[maxn];
bool done[maxn];
vector edges;
vector<int> G[maxn];

void dijkstra(int s){
    priority_queue Q;
    //_for(i,0,n) d[i] = INF;
    d[s] = 0;
    Q.push((HeapNode){0,s});   // 高端操作 
    while(!Q.empty()){
        HeapNode x = Q.top();  Q.pop();
        int u = x.u;
        if(done[u]) continue;
        done[u] = true;
        _for(i,0,G[u].size()){
            edge& e = edges[G[u][i]];
            if (d[e.v] > d[u] + e.w){
                d[e.v] = d[u] + e.w;
                Q.push((HeapNode){d[e.v], e.v});
            }
        }
    }
}

int main(){
    scanf("%d%d%d",&n,&m,&start);
    int u,v,w;
    _for(i,0,m){
        scanf("%d%d%d",&u,&v,&w);
        G[u].push_back(edges.size());
        edges.push_back(edge(u,v,w));
    }
    _rep(i,1,n) d[i] = 2147483647;
    dijkstra(start);
    _rep(i,1,n){
        printf("%d ",d[i]);
    }
    printf("\n");

    return 0;
}

bellman-ford

// O(nm)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;

const int INF = 2147483647;
const int maxn = 100000+10;
const int maxm = 500000+10;

struct edge{
    int u,v,w;
    edge(int u, int v, int w):u(u), v(v), w(w){};
};

int n,m,start,u[maxm],v[maxm],w[maxm],d[maxn];

int main(){
    scanf("%d%d%d",&n,&m,&start);
    int uu,vv,ww;
    _for(i,0,m){
        scanf("%d%d%d",&uu,&vv,&ww);
        u[i] = uu; v[i] = vv; w[i] = ww;
    }
    _rep(i,1,n) d[i] = INF;
    d[start] = 0;

    _for(k,0,n-1)   // 迭代n-1次
        _for(i,0,m){   // 检查每条边 
            int x = u[i], y = v[i];
            if( d[x] < INF ) d[y] = min(d[y], d[x]+w[i]);
        }

    _rep(i,1,n){
        printf("%d ",d[i]);
    }
    printf("\n");

    return 0;
}

bellman-ford + FIFO queue

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;

const int INF = 2147483647;
const int maxn = 100000+10;
const int maxm = 500000+10;

struct edge{
    int u,v,w;
    edge(int u, int v, int w):u(u), v(v), w(w){};
};

int n,m,start, d[maxn], cnt[maxn];
bool inq[maxn];
vector edges;
vector<int> G[maxn];

bool bellman_ford(int s){
    queue<int> Q;
    inq[s] = true;
    Q.push(s);

    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        inq[u] = false;
        _for(i,0,G[u].size()){
            edge &e = edges[G[u][i]];
            if(d[u] < INF && d[e.v] > d[u] + e.w){
                d[e.v] = d[u] + e.w;
                if(!inq[e.v]) {
                    Q.push(e.v);
                    inq[e.v] = true;
                    if (++cnt[e.v] > n) return false;   // 发现负环时及时退出? 
                }
            }
        }
    }
    return true;
}

int main(){
    scanf("%d%d%d",&n,&m,&start);
    int u,v,w;
    _for(i,0,m){
        scanf("%d%d%d",&u,&v,&w);
        G[u].push_back(edges.size());
        edges.push_back(edge(u,v,w));
    }
    _rep(i,1,n) d[i] = INF;
    d[start] = 0;

    bellman_ford(start);

    _rep(i,1,n){
        printf("%d ",d[i]);
    }
    printf("\n");

    return 0;
}

floyd
(注意不能有自边)

#include 
#include 
#include 
#include 
#include 
#define _for(i,a,b) for(int i = (a); i<(b); i++)
#define _rep(i,a,b) for(int i = (a); i<=(b); i++)
using namespace std;

const int INF = 2147483647;
const int maxn = 2000+10;
int n, m, start, d[maxn][maxn];

int main(){
    scanf("%d%d%d",&n,&m,&start);
    int u,v,w;
    _rep(i,1,n) _rep(j,1,n) d[i][j] = INF;
    _rep(i,1,n) d[i][i] = 0;
    _for(i,0,m){
        scanf("%d%d%d",&u,&v,&w);
        if (u == v) continue;  // floyd需要排除自边
        if (d[u][v] < w) continue;  // 用矩阵表示的图还要防止重边 
        d[u][v] = w;
    }

    _rep(k,1,n)
        _rep(i,1,n)
            _rep(j,1,n)
                if(d[i][k] < INF && d[k][j] < INF)
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

    _rep(i,1,n) printf("%d ",d[start][i]);
    printf("\n");

    return 0;
} 

其它

好久不见

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