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最新推荐文章于 2020-12-12 17:12:30 发布

转载最新推荐文章于 2020-12-12 17:12:30 发布 · 81 阅读

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原文链接：http://www.cnblogs.com/water-full/p/4510002.html

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#java

本文探讨了2100年海平面上升导致城市消失后的重建问题，政府计划通过建设道路来重新连接这些城市，但预算有限。文章详细介绍了输入输出格式、数据范围，并提供了一个示例输入输出，帮助读者理解如何解决这一复杂问题。

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12557 Accepted Submission(s): 3453

Problem Description

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

Author

dandelion

有些边已经联通，有些边还未联通。。。

不知道为嘛超时啊啊啊啊啊啊啊啊啊啊啊啊啊

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<vector>
using namespace std;
struct node
{
    int x,y,w;
}e[25005];
int n,m,k,cnt,fa[25005];
/*
int find(int x)
{
    if(x!=fa[x])
        fa[x]=find(fa[x]);
    return fa[x];
}
*/
int find(int x)
{
    while(x!=fa[x])
        x=fa[x];
    return fa[x];
}
int Union(int x,int y)
{
    if(x==-1)
        return 0;
    int fx,fy;
    bool flag=false;
    fx=find(x),fy=find(y);
    if(fx!=fy)
    {
        --cnt;
        fa[fx]=fy;
        flag=1;
    }
    return flag;
}
bool cmp(node a,node b)
{
    return a.w<b.w;
}
int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        int ans=0;
        scanf("%d%d%d",&n,&m,&k);
        cnt=n-1;
        for(int i=0;i<=n;i++)
            fa[i]=i;
        for(int i=1;i<=m;i++)
            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
        for(int i=0;i<k;i++)
        {
            int a,c,pre=-1;
            scanf("%d",&a);
            while(a--)
            {
                scanf("%d",&c);
                Union(pre,c);
                pre=c;
            }
        }
        sort(e+1,e+1+m,cmp);
        for(int i=1;i<=m;i++)
        {
            if(Union(e[i].x,e[i].y)==1)
                ans+=e[i].w;
        }
        if(cnt!=0)
            printf("-1\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/water-full/p/4510002.html