URAL - 1416 Confidential (最小生成树与次小生成树)

题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=12614

本文链接:http://www.cnblogs.com/Ash-ly/p/5495851.html

题意:

　　给你N个点,以及M条边,让你计算是否存在最小生成树和次小生成树,如果存在打印出权值,否则打印-1.

思路:

　　很直接的一道题,关于求次小生成树的方法,在我的另外一篇文章中:http://www.cnblogs.com/Ash-ly/p/5494975.html

代码:

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <algorithm>
  5 #include <cstdio>
  6 #include <string>
  7
  8 using namespace std;
  9 typedef long long LL;
 10
 11 const int MAXN = 500;
 12 const int MAXE = 500 * 500;
 13 const int INF = 0x3f3f3f3f;
 14 int pre[MAXN + 7];
 15
 16 void initPre(int n){ for(int i = 0; i <= n; i++) pre[i] = i; }
 17
 18 //并查集
 19 int Find(int x){ return x == pre[x] ? x : pre[x] = Find(pre[x]); }
 20
 21 void merge(int x, int y){ int fx = Find(x), fy = Find(y); if(fx != fy) pre[fx] = fy; }
 22
 23 struct Edge{ //前向星存边
 24     int u, v; //起点  终点
 25     int w;
 26     bool select;
 27 }edge[MAXE + 7];
 28
 29 bool cmp(Edge a, Edge b){
 30     if(a.w != b.w) return a.w < b.w;
 31     if(a.u != b.u) return a.u < b.u;
 32     return a.v < b.v;
 33 }
 34
 35 struct Node{//链式前向星 用于存储每个集合里面的边
 36     int to;
 37     int next;
 38 }link[MAXN + 7];
 39
 40 int head[MAXN + 7];//邻接表的头结点的位置
 41 int End[MAXN + 7];//邻接表的尾节点的位置
 42 int length[MAXN + 7][MAXN + 7];//最小生成树中任意两点路径上的最长边
 43
 44 int kruskal(int n, int m){
 45     //初始化邻接表,对于每一个顶点添加一个指向自身的边,表示以i为代表元的集合中只有点i
 46     for(int i = 1; i <= n; i++){
 47         link[i].to = i, link[i].next = head[i];
 48         End[i] = i, head[i] = i;
 49     }
 50     sort(edge + 1, edge + 1 + m, cmp);
 51     int cnt = 0;
 52     for(int i = 1; i <= m; i++){
 53         if(cnt == n - 1) break;//当找到的边数等于节点数-1,说明mst已经找到
 54         int fx = Find(edge[i].u);
 55         int fy = Find(edge[i].v);
 56         if(fx != fy){
 57             for(int j = head[fx]; j != -1; j = link[j].next)//修改length数组
 58                 for(int k = head[fy]; k != -1; k = link[k].next)
 59                 //每次合并两个等价类的之后,分别属于两个等价类的两个节点之间的最长边一定是当前加入的边
 60                     length[link[j].to][link[k].to] = length[link[k].to][link[j].to] = edge[i].w;
 61             //合并邻接表
 62             link[End[fy]].next = head[fx];
 63             End[fy] = End[fx];
 64             merge(fx, fy);
 65             cnt++;
 66             edge[i].select = true;
 67         }
 68     }
 69     if(cnt < n - 1) return -1;
 70     return 1;
 71 }
 72
 73 void init(){
 74     memset(length, -1, sizeof(length));
 75     memset(head, -1, sizeof(head));
 76     memset(&edge, 0, sizeof(Edge));
 77     memset(End, 0, sizeof(End));
 78     memset(&link, 0, sizeof(Node));
 79 }
 80
 81 int main(){
 82     //freopen("input.txt", "r", stdin);
 83     int n, m;
 84     while(~scanf("%d%d", &n, &m)){
 85         init();
 86         initPre(n);
 87         for(int i = 1; i <= m; i++) edge[i].select = false;
 88         for(int i = 1; i <= m; i++) scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
 89         int flag = kruskal(n, m);
 90         int mst = 0;
 91         for(int i = 1; i <= m; i++) if(edge[i].select) mst += edge[i].w;//计算出最小生成树
 92         int secmst = INF;
 93         //在 T/(u,v) + (x, y)中寻得次小生成树
 94         for(int i = 1; i <= m; i++) if(!edge[i].select) secmst = min(secmst, mst + edge[i].w - length[edge[i].u][edge[i].v]);
 95         if(flag < 0) printf("Cost: -1\nCost: -1\n");
 96         else if(n == m + 1) printf("Cost: %d\nCost: -1\n", mst);
 97         else printf("Cost: %d\nCost: %d\n", mst, secmst);
 98     }
 99     return 0;
100 }