第k短路

Posted 2021-11-21 ak-ls

最短路，即第1短路有很多种求法，SPFA，Dijkstra等，但第k短路怎么求呢？其实也是基于Dijkstra；因为Dijkstra用的是堆优化，这样保证每次弹出来的都是最小值，只是求最短路只是弹出一次就返回了，我们可以用Dijkstra弹出k个距离后再返回，这样根据弹出的先后顺序能够求出1~k短路

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int MAXN = 5e5 + 100;
const int MAXM = 3e3 + 10;

inline int read() {
    int x = 0, ff = 1; char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == ‘-‘) ff = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * ff;
}

inline void write(ll x) {
    if(x < 0) putchar(‘-‘), x = -x;
    if(x > 9) write(x / 10);
    putchar(x % 10 + ‘0‘);
}

int n, m, k, v;
ll ans, dist[110];
int lin[MAXN], tot = 0;
struct edge {
    int y, v, next;
}e[MAXN];

inline void add(int xx, int yy, int vv) {
    e[++tot].y = yy;
    e[tot].v = vv;
    e[tot].next = lin[xx];
    lin[xx] = tot;
}

void Dijkstra() {
    priority_queue < pair < int , int > > q;
    q.push(make_pair(0, 1));
    while(!q.empty()) {
        int x = q.top().second;
        int d = -q.top().first;
        q.pop();    
        if(x == n)  {
            dist[++v] = d;
            if(v == k + 1) return ;
        }
        for(int i = lin[x], y; i; i = e[i].next) {    
            y = e[i].y;
            ans = d + e[i].v;
            q.push(make_pair(-ans, y));
        }
    }
}

int main() {
    memset(dist, -1, sizeof(dist));
    n = read(); m = read(); k = read();
    for(int i = 1; i <= m; ++i) {
        int x,y,v;
        x = read(); y = read(); v = read();
        add(x, y, v);
    }
    Dijkstra();
    for(int i = 1; i <= k; ++i) {
        write(dist[i]);
        putchar(‘\n‘);
    }
    return 0;
}