文章标题 HDU 2122： Ice_cream’s world III （最小生成树+kruskal）

最新推荐文章于 2019-10-18 20:16:09 发布

原创最新推荐文章于 2019-10-18 20:16:09 发布 · 323 阅读

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本文详细解析了如何使用最小生成树算法解决特定问题，通过构建并查集来寻找最优路径，确保所有城市间的连接成本最低，并附带完整代码实现。

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Ice_cream’s world III

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
10

impossible

题意：就是裸的最小生成树
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;

int fa[1005];//并查集 
int n,m; 

struct node {
    int u,v,w;
    bool operator <(const node &t)const {//优先级 
        return w<t.w;
    }
}edge[10005];

int find(int x){//并查集 
    return fa[x]==x?x:fa[x]=find(fa[x]);
}

int main()
{
    while (scanf ("%d%d",&n,&m)!=EOF){
        for (int i=0;i<=n;i++){//初始化 
            fa[i]=i;
        }
        int u,v,w;
        for (int i=0;i<m;i++){//加边 
            scanf ("%d%d%d",&u,&v,&w);
            edge[i]=(node){u,v,w};
        } 
        sort(edge,edge+m);//排序 
        int sum=0;
        int ans=0;
        for (int i=0;i<m;i++){
            int fu=find(edge[i].u);
            int fv=find(edge[i].v);
            if (fu!=fv){//不在同一个联通快则加入最小生成树 
                sum++;//生成树的边数加1 
                fa[fu]=fv;
                ans+=edge[i].w;
            }
            if (sum==n-1)break;
        }
        if (sum==n-1){
            printf ("%d\n\n",ans);
        }else printf ("impossible\n\n");
    }
    return 0;
}