CodeForces 437C The Child and Toy

最新推荐文章于 2017-08-07 21:45:43 发布

peach_yang

最新推荐文章于 2017-08-07 21:45:43 发布

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本文介绍了一个关于玩具拆分的问题，该问题需要计算拆分玩具所需的最小总能量。通过使用贪心算法，文章详细阐述了如何对玩具各部分的能量进行排序，并在两部分由绳子连接时选择能量较低的部分进行拆分，从而达到最小化总能量消耗的目的。

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Question：
On Children’s Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can’t wait to destroy the toy.

The toy consists of n parts and m ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let’s define an energy value of part i as vi. The child spend vf1 + vf2 + … + vfk energy for removing part i where f1, f2, …, fk are the parts that are directly connected to the i-th and haven’t been removed.

Help the child to find out, what is the minimum total energy he should spend to remove all n parts.

Input
The first line contains two integers n and m (1 ≤ n ≤ 1000; 0 ≤ m ≤ 2000). The second line contains n integers: v1, v2, …, vn (0 ≤ vi ≤ 105). Then followed m lines, each line contains two integers xi and yi, representing a rope from part xi to part yi (1 ≤ xi, yi ≤ n; xi ≠ yi).

Consider all the parts are numbered from 1 to n.

Output
Output the minimum total energy the child should spend to remove all n parts of the toy.

Sample Input
Input
4 3
10 20 30 40
1 4
1 2
2 3
Output
40
Input
4 4
100 100 100 100
1 2
2 3
2 4
3 4
Output
400
Input
7 10
40 10 20 10 20 80 40
1 5
4 7
4 5
5 2
5 7
6 4
1 6
1 3
4 3
1 4
Output
160
题意大意：一个人要把玩具拆分至不可拆分的部分，拆下每个部分都会消耗一部分能量，并且每两部分至少有一个绳子连接。问拆下这下玩具至少消耗多少能量。
思路：这道题可用贪心来算，将拆下消耗的能量从大到小排序，如果两个部分用绳子连在一起就选取能量较小的那部分加上即可。
(http://acm.hust.edu.cn/vjudge/contest/123566#problem/R)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
    int eng;
    int num;
}a[1005];
int n,m,f[1005][1005];
bool cmp(node x,node y)
{
    return x.eng>y.eng;
}
int main()
{
    memset(f,0,sizeof(f));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].eng);
        a[i].num=i;
    }
    for(int i=0;i<m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        f[x][y]=f[y][x]=1;   //处理两部分用绳子连在一起
    }
    sort(a+1,a+1+n,cmp);
    int total=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(f[a[i].num][a[j].num]==1)     //如果两部分用绳子连在一起，则选取能量较少的那部分加上
            {
                total+=a[j].eng;
                f[a[i].num][a[j].num]=f[a[j].num][a[i].num]=0;
            }
        }
    }
    printf("%d\n",total);
   return 0;
}