Bad Cowtractors

Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

Output

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

Sample Input

5 81 2 31 3 72 3 102 4 42 5 83 4 63 5 24 5 17

Sample Output

42

题解：克鲁斯卡尔不用考虑重边。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int INF = 0x3fffffff;struct Node{int from;int to;int cost;bool operator< (Node t) const{return cost > t.cost;}};Node e[200004];int pre[20000];int ans;int find(int x){return x == pre[x] ? x : pre[x] = find(pre[x]);}int max(int a,int b){return a > b ? a : b;}bool kruaskal(int n,int k){int j = 0;ans = 0;int cnt = 0;for(int i = 0;i < n;i++){for(;j < k;j++){int x = find(e[j].from);int y = find(e[j].to);if(x == y){continue;}++cnt;pre[x] = y;ans += e[j].cost;break;}}if(cnt != n - 1){return false;}return true;}int main(){int n,m;while(scanf("%d%d",&n,&m) != EOF){for(int i = 1;i <= n;i++){pre[i] = i;}int k = 0;for(int i = 0;i < m;i++){int u,v,c;scanf("%d%d%d",&u,&v,&c);e[k].from = u;e[k].to = v;e[k++].cost = c;e[k].from = v;e[k].to = u;e[k++].cost = c;}sort(e,e + k);if(kruaskal(n,k))    printf("%d\n",ans);else    printf("-1\n");}return 0;}

普利姆（未优化）：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int INF = 0x3fffffff;int map[2003][2003];bool visited[2003];int d[2003];int ans;int max(int a,int b){return a > b ? a : b;}bool prim(int n){memset(visited,false,sizeof(visited));for(int i = 1;i <= n;i++){d[i] = map[1][i];}visited[1] = true;ans = 0;for(int i = 1;i < n;i++){int min = 0;int k;for(int j = 1;j <= n;j++){if(!visited[j] && min < d[j]){min = d[j];k = j;}}if(min == 0){return false;}ans += min;visited[k] = true;for(int j = 1;j <= n;j++){if(!visited[j] && d[j] < map[k][j]){d[j] = map[k][j];}}}return true;}int main(){int n,m;while(scanf("%d%d",&n,&m) != EOF){for(int i = 1;i <= n;i++){for(int j = 1;j <= n;j++){map[i][j] = 0;}}int u,v,c;for(int i = 0;i < m;i++){scanf("%d%d%d",&u,&v,&c);map[u][v] = map[v][u] = max(map[u][v],c);}if(prim(n)){printf("%d\n",ans);}else{printf("-1\n");}}return 0;}