Connect the Cities

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  

Input

The first line contains the number of test cases. 
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities. 
To make it easy, the cities are signed from 1 to n. 
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q. 
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities. 

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

Sample Output

1

C++版本一

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
const int N=510;
int T,t,m,n,k,pre[N],id[N];
struct node{
    int f,t,w;
    node(){};
    node(int a,int b,int c){
        f=a;
        t=b;
        w=c;
    }
    bool operator < (const node &S )const{

        return w<S.w;
    }
}e[N*N];
int find(int x){
    int r=x;
    while(pre[r]!=r)
        r=pre[r];
    return r;
}
void join(int x,int y){
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
        pre[fx]=fy;
}
int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m,&k);
        int a,b,c;
        int cnt=0;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            e[++cnt]=node(a,b,c);
        }
        sort(e+1,e+cnt+1);
        for(int i=1;i<=n;i++)
            pre[i]=i;
        for(int K=1;K<=k;K++){
            scanf("%d",&t);
            for(int i=1;i<=t;i++){
                scanf("%d",&id[i]);
                for(int j=1;j<i;j++){
                    join(id[i],id[j]);
                }
            }
        }
        int ans=0;
        for(int i=1;i<=cnt;i++)
            if(find(e[i].f)!=find(e[i].t)){
                join(e[i].f,e[i].t);
                ans+=e[i].w;
            }
        int flag=0;
        for(int i=1;i<=n;i++)
            if (pre[i]==i)
                flag++;
        if(flag==1)
            cout << ans << endl;
        else
            cout << -1 << endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <climits>

using namespace std;

const int MAX = 25003;

typedef struct Road{
	int x;
	int y;
	int c;
}Road;

Road road[MAX];
int cr;

int pre[501];

void init(int n){
	int i;
	for(i=1;i<=n;++i){
		pre[i] = i;
	}
	cr = n-1;
}

int root(int x){
	if(x!=pre[x]){
		pre[x] = root(pre[x]);
	}
	return pre[x];
}

int merge(int x,int y){
	if(x==-1)return 0;
	int ret = 0;
	int fx = root(x);
	int fy = root(y);
	if(fx!=fy){
		--cr;
		pre[fx] = fy;
		ret = 1;
	}
	return ret;
}

int cmp(const void *a,const void *b){
	return ((Road *)a)->c - ((Road *)b)->c;
}

int main(){
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	int t,n,m,k,cid,tc,pre;
	int sum,i,j;
	scanf("%d",&t);
	while(t--){
		scanf("%d %d %d",&n,&m,&k);
		init(n);
		for(i=0;i<m;++i){
			scanf("%d %d %d",&road[i].x,&road[i].y,&road[i].c);
		}
		sum = 0;
		for(i=0;i<k;++i){
			scanf("%d",&tc);
			pre = -1;
			for(j=0;j<tc;++j){
				scanf("%d",&cid);
				merge(pre,cid);
				pre = cid;
			}
		}
		qsort(road,m,sizeof(Road),cmp);
		for(i=0;i<m;++i){
			if(merge(road[i].x,road[i].y)==1){
				sum += road[i].c;
			}
		}
		if(cr!=0){
			printf("-1\n");
		}else{
			printf("%d\n",sum);
		}
	}
    return 0;
}

C++版本三

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <climits>

using namespace std;

const int MAX = 25003;

typedef struct Road{
	int x;
	int y;
	int c;
}Road;

Road road[MAX];
int cr;

int pre[501];

void init(int n){
	int i;
	for(i=1;i<=n;++i){
		pre[i] = i;
	}
	cr = n-1;
}

int root(int x){
	if(x!=pre[x]){
		pre[x] = root(pre[x]);
	}
	return pre[x];
}

int merge(int x,int y){
	if(x==-1)return 0;
	int ret = 0;
	int fx = root(x);
	int fy = root(y);
	if(fx!=fy){
		--cr;
		pre[fx] = fy;
		ret = 1;
	}
	return ret;
}

int cmp(const void *a,const void *b){
	return ((Road *)a)->c - ((Road *)b)->c;
}

int main(){
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	int t,n,m,k,cid,tc,pre;
	int sum,i,j;
	scanf("%d",&t);
	while(t--){
		scanf("%d %d %d",&n,&m,&k);
		init(n);
		for(i=0;i<m;++i){
			scanf("%d %d %d",&road[i].x,&road[i].y,&road[i].c);
		}
		sum = 0;
		for(i=0;i<k;++i){
			scanf("%d",&tc);
			pre = -1;
			for(j=0;j<tc;++j){
				scanf("%d",&cid);
				merge(pre,cid);
				pre = cid;
			}
		}
		qsort(road,m,sizeof(Road),cmp);
		for(i=0;i<m;++i){
			if(merge(road[i].x,road[i].y)==1){
				sum += road[i].c;
			}
		}
		if(cr!=0){
			printf("-1\n");
		}else{
			printf("%d\n",sum);
		}
	}
    return 0;
}

 

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03-03 23:12
已编辑
北京邮电大学 Java
书海为家:我来给一点点小建议,因为毕竟还在学校不像工作几年的老鸟有丰富的项目经验,面试官在面试在校生的时候更关注咱们同学的做事逻辑和思路,所以最好在简历中描述下自己做过项目的完整过程,比如需求怎么来的,你对需求的解读,你想到的解决办法,遇到困难如何找人求助,最终项目做成了什么程度,你从中收获了哪些技能,你有什么感悟。
你的简历改到第几版了
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