Johnson算法

求多源负权最短路时用，比\(floyd\)快，\(O(VE + V^2logV)\)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a,b,sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e_Line printf("\n---------------\n")
#define D_e(x) cout << (#x) << " : " << x << "\n"
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#include <ctime>
#define TIME() fprintf(stderr, "TIME : %.3lfms\n", (double)clock() * 1.0 / (double)CLOCKS_PER_SEC)
#endif
struct FastIO {
    template<typename ATP> inline FastIO& operator >> (ATP &x) {
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        if(f == -1) x = -x;
        return *this;
    }
} io;
using namespace std;
template<typename ATP> inline ATP Max(ATP x, ATP y) {
    return x > y ? x : y;
}
template<typename ATP> inline ATP Min(ATP x, ATP y) {
    return x < y ? x : y;
}

const int N = 5007;

struct Edge {
    int nxt, pre, w, from;
} e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v, int w) {
    e[++cntEdge] = (Edge){ head[u], v, w, u}, head[u] = cntEdge;
}

int DIS[N][N], H[N], vis[N];
int qq[N], h, t;
int n, m;
inline void SPFA(int st) {
    for(register int i = 1; i <= n; i += 3){
        H[i] = 0x3f3f3f3f;
        H[i + 1] = 0x3f3f3f3f;
        H[i + 2] = 0x3f3f3f3f;
    }
    H[st] = 0;
    qq[++t] = st;
    while(h != t){
        int u = qq[++h];
        if(h >= N - 5) h = 0;
        vis[u] = false;
        for(register int i = head[u]; i; i = e[i].nxt){
            int v = e[i].pre;
            if(H[v] > H[u] + e[i].w){
                H[v] = H[u] + e[i].w;
                if(!vis[v]){
                    vis[v] = true;
                    qq[++t] = v;
                    if(t >= N - 5) t = 0;
                }
            }
        }
    }
}

struct nod {
    int x, w;
    bool operator < (const nod &com) const {
        return w > com.w;
    }
};
#include <queue>
priority_queue<nod> q;
int dis[N];
inline void Dijkstra(int st) {
    for(register int i = 1; i <= n; i += 3){
        dis[i] = 0x3f3f3f3f;
        dis[i + 1] = 0x3f3f3f3f;
        dis[i + 2] = 0x3f3f3f3f;
    }
    dis[st] = 0;
    q.push((nod){ st, 0});
    while(!q.empty()){
        int u = q.top().x, w = q.top().w;
        q.pop();
        if(w != dis[u]) continue;
        for(register int i = head[u]; i; i = e[i].nxt){
            int v = e[i].pre;
            if(dis[v] > dis[u] + e[i].w){
                dis[v] = dis[u] + e[i].w;
                q.push((nod){ v, dis[v]});
            }
        }
    }
}

int main() {
    io >> n >> m;
    
    R(i,1,m){
        int u, v, w;
        io >> u >> v >> w;
        add(u, v, w);
    }
    
    R(i,1,n){
        add(0, i, 0);
    }
    
    SPFA(0);

    R(i,1,cntEdge){
        e[i].w += H[e[i].from] - H[e[i].pre];
    }

    R(i,1,n){
        Dijkstra(i);
        R(j,1,n){
            DIS[i][j] = dis[j] - H[i] + H[j];
        }
    }
    
    R(i,1,n){
        R(j,1,n){
            printf("%d ", DIS[i][j]);
        }
        putchar('\n');
    }
    
    return 0;
}
/*
5 5
1 2 9
1 4 7
2 4 -3
2 3 11
5 2 -2
*/

标签：return,Johnson,int,dis,算法,ATP,include,define
来源： https://www.cnblogs.com/bingoyes/p/11765859.html