Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 575 Accepted Submission(s): 184

Problem Description

ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.

Input

Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.

Output

If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.

Sample Input

2 1

0 1 10

4 0

Sample Output

impossible

Author

Wiskey

Source

HDU 2007-10 Programming Contest_WarmUp

Recommend

威士忌

思路：Kruskal

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 10060;
int map[1100];
int n,m;
int the_last_flag;
int a,b,how_long;
struct Node
{int first,end;int value;int select;
}Edge[MAXN];
int find(int x)
{while(x != map[x]){x = map[x];}return x;
}
void Merge(int x,int y)
{int p = find(x);int q = find(y);if(p < q)map[q] = p;elsemap[p] = q;
}
bool cmp(Node a,Node b)
{if(a.value != b.value)return a.value < b.value;if(a.first != b.first)return a.first < b.first;return a.end < b.end;
}
void Kruskal(Node *Edge,int n,int m)
{the_last_flag = 0;sort(Edge + 1,Edge + m + 1,cmp);for(int i = 1;i <= m;i ++){if(the_last_flag == n - 1)break ;int x = find(Edge[i].first);int y = find(Edge[i].end);if(x != y){Merge(x,y);Edge[i].select = true;the_last_flag ++;}}
}
int main()
{while(~scanf("%d%d",&n,&m)){for(int i = 1;i <= n;i ++)map[i] = i;memset(Edge,0,sizeof(Edge));for(int i = 1;i <= m;i ++){scanf("%d%d%d",&a,&b,&how_long);a ++;b ++;Edge[i].first = a;Edge[i].end = b;Edge[i].value = how_long;}Kruskal(Edge,n,m);if(the_last_flag != n - 1)printf("impossible\n");else{int the_last_sum = 0;for(int i = 1;i <= m;i ++){if(Edge[i].select == 1)the_last_sum += Edge[i].value;}printf("%d\n",the_last_sum);}printf("\n");}return 0;
}

转载于:https://www.cnblogs.com/GODLIKEING/p/3376849.html

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