图论相关算法

转载自：http://cojs.tk/cogs/page/page.php?aid=30

 

 

    • 最小生成树
      • Kruskal+ufs
      •  1 int ufs(int x){
 2     return f[x] == x ? x : f[x] = ufs(f[x]); 
 3 } 
 4 int Kruskal(){
 5    int w = 0;
 6    for(int i=0; i<n; i++)
 7        f[i] = i;
 8    sort(e, e+n);
 9    for(int i=0; i<n; i++){
10        int x = ufs(e[i].u), y = ufs(e[i].v);
11        if(x != y) {
12            f[x] = y;
13            w += e[i].w;
14        }
15    } 
16    return w; 
17 }

         

      • Prim 
         1 int Prim() {
 2        int w = 0;
 3        priority_queue<pair<int, int> > q;
 4        bool l[N] = {0};
 5        l[1] = 1; q.push(make_pair(0, 1));
 6        for(int k=1; k<n; k++) {
 7            int u = q.top().second; q.pop();
 8            for(int i=0; i<G[u].size(); i++)
 9                if(!l[G[u][i]])
10                    q.push(make_pair(-c[u][i], G[u][i]));
11            while(!q.empty() && l[q.top().second])
12                q.pop();
13            l[q.top().second] = 1;
14            w += -q.top().first;
15            q.pop();
16        } return w; }
  • 最短路径
    •  Dijkstra+priority_queue 
       1 void Dijkstra(int s) {
 2        priority_queue<pair<int, int> > q;
 3        bool l[N] = {0}; l[s] = 1;
 4        fill_n(f, n, INF); f[s] = 0;
 5        q.push(make_pair(-f[s], s));
 6        while(!q.empty()) {
 7            int u = q.front().second; q.pop();
 8            for(int i=0; i<G[u].size(); i++) {
 9                int v = G[u][i];
10                if(f[v] > f[u] + c[u][i]) {
11                    f[v] = f[u] + c[u][i];
12                    if(!l[v]) {
13                        l[v] = 1;
14                        q.push(make_pair(-f[v], v));
15                    }
16                }
17            }
18        } }
    • Bellman-Ford (SPFA) 
       1 void BellmanFord(int s) { // SPFA
 2        queue<int> q;
 3        bool l[N] = {0}; l[s] = 1;
 4        fill_n(f, n, INF); f[s] = 0;
 5        q.push(s);
 6        while(!q.empty()) {
 7            int u = q.front(); q.pop();
 8            l[u] = 0;
 9            for(int i=0; i<G[u].size(); i++) {
10                int v = G[u][i];
11                if(f[v] > f[u] + c[u][i]) {
12                    f[v] = f[u] + c[u][i];
13                    if(!l[v]) {
14                        l[v] = 1;
15                        q.push(v);
16                    }
17                }
18            }
19        } }
    • Floyd 
      1 void Floyd() {
2        for(int k=0; k<n; k++)
3            for(int i=0; i<n; i++)
4                for(int j=0; j<n; j++)
5                    f[i][j] = min(f[i][j], f[i][k] + f[k][j]); }

       

  • 二分图
    • ufs 验证 Hungary 
       1 bool DFS(int u) {
 2        for(int i=0; i<G[u].size(); i++) {
 3            int v = G[u][i];
 4            if(!l[v]) {
 5                l[v] = 1;
 6                if(!f[v] || DFS(f[v])) {
 7                    f[v] = u;
 8                    return true;
 9                }
10            }
11        } return false; } int Hungary() {
12        int w = 0;
13        for(int i=0; i<n; i++) {
14            fill_n(l, l+n, 0);
15            if(DFS(i))
16                w++;
17        } return w; }
  • 连通分量
    • Tarjan stack 
       1 <int> s; void Tarjan(int u) {
 2        dfn[u] = low[u] = ++time;
 3        l[u] = 1;
 4        s.push(u);
 5        for(int i=0; i<G[u].size(); i++) {
 6            int v = G[u][i];
 7            if(!dfn[v]) {
 8                Tarjan(v);
 9                low[u] = min(low[u], low[v]);
10            } else if(l[v])
11                low[u] = min(low[u], dfn[v]);
12        }
13        if(dfn[u] == low[u]) {
14            w++;
15            do {int v;
16                l[v = s.top()] = 0;
17                f[v] = w;
18                s.pop();
19            } while(u != v);
20        } } void SCC() {
21        fill_n(dfn, n, 0);
22        for(int i=0; i<n; i++)
23            if(!dfn(i))
24                Tarjan(i); }

       

  • 网络流
    • 费用流：Bellman-Ford 找增广路，或者用贪心求解
    • 最大流：Edmonds-Karp
posted @ 2015-07-27 18:20  OhYee  阅读(253)  评论(0)  编辑  收藏  举报