[POJ2377]Bad Cowtractors(最大生成树，Kruskal)

题目链接：http://poj.org/problem?id=2377

于是就找了一道最大生成树的AC了一下，注意不连通的情况啊，WA了一次。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%lld", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef map<string, int> msi;
 65 typedef vector<int> vi;
 66 typedef vector<LL> vl;
 67 typedef vector<vl> vvl;
 68 typedef vector<bool> vb;
 69 
 70 typedef struct Edge {
 71     int u, v, w;
 72     Edge() {}
 73     Edge(int uu, int vv, int ww) : u(uu), v(vv), w(ww) {}
 74 }Edge;
 75 const int maxn = 1010;
 76 const int maxm = 20020;
 77 int n, m;
 78 int pre[maxn];
 79 Edge edge[maxm];
 80 
 81 bool cmp(Edge a, Edge b) {
 82     RT a.w > b.w;
 83 }
 84 
 85 int find(int x) {
 86     return x == pre[x] ? x : pre[x] = find(pre[x]);
 87 }
 88 
 89 int unite(int x, int y) {
 90     int fx = find(x);
 91     int fy = find(y);
 92     if(fx != fy) {
 93         pre[fy] = fx;
 94         return 1;
 95     }
 96     return 0;
 97 }
 98 
 99 int main() {
100     // FRead();
101     int u, v, c;
102     while(~Rint(n) && ~Rint(m)) {
103         Rep(i, n+5) pre[i] = i;
104         Rep(i, m) {
105             Rint(u); Rint(v); Rint(c);
106             edge[i] = Edge(u, v, c);
107         }
108         sort(edge, edge+m, cmp);
109         int ret = 0;
110         Rep(i, m) {
111             if(unite(edge[i].u, edge[i].v)) {
112                 ret += edge[i].w;
113             }
114         }
115         int flag = 0;
116         For(i, 1, n+1) {
117             if(pre[i] == i) flag++;
118             if(flag > 1) break;
119         }
120         if(flag > 1) printf("-1\n");
121         else printf("%d\n", ret);
122     }
123     RT 0;
124 }

 

posted @ 2016-05-31 11:11  Kirai  阅读(150)  评论(0)  编辑  收藏  举报