HDU 畅通工程系列

畅通工程系列都是比较裸的最小生成树问题，且是中文题目，不赘述了。

1、HDU 1863 畅通工程

题意：一个省有很多村庄，其中一些之间是可以建公路的，每条公路都需要不同的代价，问代价最小的情况下将所有村庄都连通。

解法：裸的最小生成树。

tag:MST

由于才学MST，两种算法一个写了一遍：

Kruskal：

 /*
  * Author:  Plumrain
  * Created Time:  2013-11-30 10:21
  * File Name: G-HDU-1863.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 typedef pair<int, int> pii;
 struct Pat{
     int s, e, w;
 };

 int n, m, f[];
 Pat p[];

 bool cmp(Pat a, Pat b)
 {
     return a.w < b.w;
 }

 void init()
 {
     for (int i = ; i < n; ++ i){
         scanf ("%d%d%d", &p[i].s, &p[i].e, &p[i].w);
         -- p[i].s; --p[i].e;
     }
 }

 int find(int x)
 {
     if (x != f[x]) f[x] = find(f[x]);
     return f[x];
 }

 int kruskal(int n, int m)
 {
     sort(p, p+m, cmp);
     for (int i = ; i < n; ++ i)
         f[i] = i;
     int cost = ;
     for (int i = ; i < m; ++ i){
         int s = p[i].s, e = p[i].e;
         int t1 = find(s), t2 = find(e);
         if (t1 != t2){
             f[t1] = t2;
             cost += p[i].w;
         }
     }

     int tmp = find();
     for (int i = ; i < n; ++ i){
         find (i);
         if (f[i] != tmp) return -;
     }
     return cost;
 }

 int main()
 {
     while (scanf ("%d%d", &n, &m) != EOF && n){
         init();
         int ans = kruskal(m, n);
         if (ans == -) printf ("?\n");
         else printf ("%d\n", ans);
     }
     return ;
 }

Prim：

 /*
  * Author:  Plumrain
  * Created Time:  2013-11-30 10:51
  * File Name: G-HDU-1863.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <utility>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define PB push_back
 const int maxint = ;
 typedef pair<int, int> pii;

 int n, m;
 bool v[];
 int c[];
 vector<pii> p[];

 struct cmp{
     bool operator() (pii a, pii b){
         return a.second > b.second;
     }
 };

 void init()
 {
     for (int i = ; i < n; ++ i)
         p[i].clear();

     int a, b, w;
     for (int i = ; i < m; ++ i){
         scanf ("%d%d%d", &a, &b, &w);
         -- a; -- b;
         p[a].PB (make_pair(b, w));
         p[b].PB (make_pair(a, w));
     }
 }

 int prim(int n, int m)
 {
     int cost = ;
     CLR (v);
     priority_queue<pii, vector<pii>, cmp> q;
     while (!q.empty()) q.pop();

     for (int i = ; i < n; ++ i)
         c[i] = maxint - ;
     c[] = ;
     v[] = ;
     for (int i = ; i < (int)p[].size(); ++ i){
         pii tmp = p[][i];
         c[tmp.first] = tmp.second;
         q.push (make_pair(tmp.first, tmp.second));
     }

     for (int i = ; i < n-; ++ i){
         if (q.empty()) break;
         pii tmp = q.top(); q.pop();
         while (v[tmp.first] && !q.empty()){
             tmp = q.top(); q.pop();
         }
         if (v[tmp.first]) break;

         int t1 = tmp.first, t2 = tmp.second;
         cost += t2;
         v[t1] = ;
         c[t1] = ;
         for (int j = ; j < (int)p[t1].size(); ++ j) if (!v[p[t1][j].first]){
             pii t = p[t1][j];
             if (c[t.first] > t.second){
                 q.push (make_pair(t.first, t.second));
                 c[t.first] = t.second;
             }
         }
     }

     for (int i = ; i < n; ++ i)
         if (!v[i]) return -;
     return cost;
 }

 int main()
 {
     while (scanf ("%d%d", &m, &n) != EOF && m){
         init();
         int ans = prim(n, m);
         if (ans == -) printf ("?\n");
         else printf ("%d\n", ans);
     }
     return ;
 }

2、HDU 1875 畅通工程再续

没写代码。

3、HDU 1879 继续畅通工程

用Kruskal写了一份：

 /*
  * Author:  Plumrain
  * Created Time:  2013-12-01 02:52
  * File Name: G-HDU-1879.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 struct Pat{
     int s, e, w;
 };

 int n, all, f[];
 Pat p[];

 bool cmp(Pat a, Pat b)
 {
     return a.w < b.w;
 }

 int find (int x)
 {
     if (x != f[x]) f[x] = find(f[x]);
     return f[x];
 }

 void init()
 {
     all = ;
     for (int i = ; i < n; ++ i)
         f[i] = i;

     int s, e, w, x;
     for (int i = ; i < n*(n-)/; ++ i){
         scanf ("%d%d%d%d", &s, &e, &w, &x);
         -- s; -- e;
         if (x == ){
             int t1 = find(s), t2 = find(e);
             if (t1 != t2) f[t1] = t2;
         }
         else{
             p[all].s = s; p[all].e = e;
             p[all++].w = w;
         }
     }
 }

 int kruskal()
 {
     int cost = ;
     sort(p, p+all, cmp);
     for (int i = ; i < all; ++ i){
         int s = p[i].s, e = p[i].e, w = p[i].w;
         int t1 = find(s), t2 = find(e);
         if (t1 != t2){
             f[t1] = t2; cost += w;
         }
     }
     return cost;
 }

 int main()
 {
     while (scanf ("%d", &n) != EOF && n){
         init();
         int ans = kruskal();
         printf ("%d\n", ans);
     }
     return ;
 }

4、HDU 1233 还是畅通工程

用Prim写了一份：

 /*
  * Author:  Plumrain
  * Created Time:  2013-12-02 01:43
  * File Name: G-HDU-1233.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <utility>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 typedef pair<int, int> pii;

 int n, m, d[];
 bool v[];
 vector<pii> pat[];

 struct cmp{
     bool operator() (pii a, pii b){
         return a.second > b.second;
     }
 };

 void init()
 {
     for (int i = ; i < n; ++ i)
         pat[i].clear();

     m = n * (n-) / ;
     int a, b, w;
     for (int i = ; i < m; ++ i){
         scanf ("%d%d%d", &a, &b, &w);
         -- a; -- b;
         pat[a].PB (make_pair(b, w));
         pat[b].PB (make_pair(a, w));
     }
 }

 int Prim(int n, int m)
 {
     priority_queue<pii, vector<pii>, cmp> q;
     while (!q.empty()) q.pop();

     int cost = ;
     CLR1 (d);
     CLR (v); v[] = ; d[] = ;
     int t_sz = pat[].size();
     for (int i = ; i < t_sz; ++ i){
         pii tmp = pat[][i];
         d[tmp.first] = tmp.second;
         q.push (tmp);
     }

     for (int i = ; i < n-; ++ i){
         if (q.empty()) break;
         pii tmp = q.top(); q.pop();
         while (v[tmp.first] && !q.empty()){
             tmp = q.top(); q.pop();
         }
         if (v[tmp.first]) break;

         int t1 = tmp.first, t2 = tmp.second;
         cost += t2; v[t1] = ; d[t1] = ;
         int sz = pat[t1].size();
         for (int j = ; j < sz; ++ j) if (!v[pat[t1][j].first]){
             pii tt = pat[t1][j];
             if (d[tt.first] > tt.second){
                 q.push (make_pair(tt.first, tt.second));
                 d[tt.first] = tt.second;
             }
         }
     }

     for (int i = ; i < n; ++ i)
         if (!v[i]) return -;
     return cost;
 }

 int main()
 {
     while (scanf ("%d", &n) != EOF && n){
         init();
         int ans = Prim(n, m);
         printf ("%d\n", ans);
     }
     return ;
 }