Floyd and Dijkstra shortest path algorithm

1.  problem

　　The shortest, as the name suggests, is to find the shortest path. Specifically, on a graph, given the starting point s and the end point t, remove the weight and minimal side, so that to reach from the s t.

2.  resolve

　　1, floyd algorithm

　　　　This algorithm is a more violent algorithm, each enumeration three points A , B , C , if the DIS (A, B)> DIS (A, C) + DIS (C, B) , then the description, A- > b paths, I can select a point C , the dis (a, b) updates the value of DIS (a, C) + DIS (C, B) , we call such an operation a relaxed, continuously in constant enumeration in slack, DIS (a, B) will represent a-> b minimum distance.

 

　　2, dijkstra algorithm

　　　　① . Initially, S contains only a starting point s ; U comprising inter s other vertex outside, and U in the distance from the apex of " starting point s to the distance from the vertex " [ e.g., U vertex v distance is (s, v ) length, and s and v are not adjacent, then v is a distance ∞] .

　　　　② . From U elect the " shortest distance vertex k" , and the vertex k is added to S ; the same time, the U remove vertices k .

　　　　③ . Update U of each point to the starting point s distance. The reason why the update U distance vertices, due to the previous step to determine the k is determined vertices shortest path, thereby using k to update the other vertex distance; such as, (s, V) the distance may be greater than (s, k) + (k, v) a distance

　　　　④ . Repeat steps (2) and (3) , until all vertices been traversed.

3.  Design

　　1, Floyd algorithm

for (int i = 1; i <= n; i++) {
　　for (int k = 1; k <= n; k++) {
　　　　for (int j = 1; j <= n; j++) {
　　　　　　p[k][j] = min(p[k][j], p[k][i] + p[i][j]);//松弛操作
　　　　}
　　}
}

 

　　2, Dijkstra algorithm

the while (queue not empty) {
　　int DIS A = team head, b = X team head;
　　for (side connected to the point b) {
　　　　IF (W + A < DIS [to]) { 
　　　　　　DIS [to] = W + A; 
　　　　　　(DIS [to], to) enqueue; 
　　　　} 
　　} 
}

 

4.  Analysis

　　1, no doubt, Floyd time complexity is O (n- ³ ) , this algorithm has advantages and disadvantages, of course, is a disadvantage of complexity is too high, the efficiency is too slow; the advantage of easy interrogation, each query complexity is O (1) . It pretreated shortest distance between each of all the points.

　　2, Dijkstra complexity easily calculated, Common complexity is O (n- ² ) , optimization point is how to remove from the set of nearest point, I used rootlets stack can O (1) taken out nearest points, update operation requires O (logN) , summary, the complexity of the stack is optimized O (nlogn) .

5. Source

 　　1、Floyd

/*
author: keke
project name:Floyd算法
Time Complexity: O(n³)
*/
#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
int mo[4][2] = { -1,0,1,0,0,-1,0,1 };
//int mo[8][2] = { -1,0,1,0,-1,-1,1,1,0,-1,0,1,1,-1,-1,1 };
const int maxn = 110;
ll n, m, is, t, ans, p[maxn][maxn], a, b, c;
int main() {
#ifdef ONLINE_JUDGE
#else
    //freopen("input.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(2);
    while (cin >> n >> m) {
        for (int i = 1; i <= n; i++) {
            for (int k = 1; k <= n; k++) {
                if (i == k)p[i][k] = 0;
                else p[i][k] = 1e18;
            }
        }
        for (int i = 1; i <= m; i++) {
            cin >> a >> b >> c;
            p[a][b] = p[b][a] = min(c, p[a][b]);//二维矩阵存图
        }
        for (int i = 1; i <= n; i++) {
            for (int k = 1; k <= n; k++) {
                for (int j = 1; j <= n; j++) {
                    p[k][j] = min(p[k][j], p[k][i] + p[i][j]);//松弛操作
                }
            }
        }
        cin >> a >> b;
        if (p[a][b] == 1e18)cout << "No path" << "\n";
        else cout << p[a][b] << "\n";
    }
    return 0;
    //good job!
}

 

　　2、Dijkstra

/*
author: keke
project name:Dijkstra算法(堆优化)
Time Complexity: O(nlog(n))
*/
#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
int mo[4][2] = { -1,0,1,0,0,-1,0,1 };
//int mo[8][2] = { -1,0,1,0,-1,-1,1,1,0,-1,0,1,1,-1,-1,1 };
const int maxn = 110;
int n, m, is, t, ans, cnt, head[maxn], a, b, c, dis[maxn];
bool vis[maxn];
struct xx {
    int t, next, w;
}xx[maxn << 1];
void add(int a, int b, int c) {//链式前向星
    xx[cnt].next = head[a];
    xx[cnt].t = b;
    xx[cnt].w = c;
    head[a] = cnt++;
}
void solve(int x, int y) {
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++)dis[i] = inf;
    priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
    while (!q.empty())q.pop();
    dis[x] = 0;
    q.push(make_pair(0, x));//采用堆优化
    while (!q.empty()) {
        int a = q.top().first, b = q.top().second;
        q.pop();
        if (vis[b])continue;
        vis[b] = true;
        for (int i = head[b]; ~i; i = xx[i].next) {
            int to = xx[i].t;
            if (xx[i].w + a < dis[to]) {//松弛操作
                dis[to] = a + xx[i].w;
                q.push(make_pair(dis[to], to));
            }
        }
    }
    if (dis[y] == inf)cout << "No path" << "\n";
    else cout << dis[y] << "\n";
}
int main() {
#ifdef ONLINE_JUDGE
#else
    //freopen("input.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(2);
    while (cin >> n >> m) {
        cnt = 0;
        memset(head, -1, sizeof(head));
        for (int i = 1; i <= m; i++) {
            cin >> a >> b >> c;
            add(a, b, c);
            add(b, a, c);
        }
        cin >> a >> b;
        solve(a, b);
    }
    return 0;
    //good job!
}