CodeForces - 437C The Child and Toy （贪心）

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于 2019-07-31 10:36:00 发布

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分类专栏：贪心 CodeForces

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儿童节时，孩子收到一个玩具，但他想破坏它。玩具由n个部件和m条绳子组成，每个绳子连接两个部件。孩子必须逐个移除所有部件，移除部件i消耗的能量为与其直接相连且未移除的部件能量之和。问题是找出移除所有部件的最小总能量。输入包括部件数n、绳子数m和每个部件的能量值，以及绳子连接的部件信息。解决这个问题的一种方法是每次都移除剩余部件中能量值最大的那一个，以确保最低能耗。

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The Child and Toy

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.

Examples

Input

4 3
10 20 30 40
1 4
1 2
2 3

Output

Input

4 4
100 100 100 100
1 2
2 3
2 4
3 4

Output

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

Note

One of the optimal sequence of actions in the first sample is:

First, remove part 3, cost of the action is 20.
Then, remove part 2, cost of the action is 10.
Next, remove part 4, cost of the action is 10.
At last, remove part 1, cost of the action is 0.

So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.

In the second sample, the child will spend 400 no matter in what order he will remove the parts.

题目链接：http://codeforces.com/problemset/problem/437/C

题目大意：n个点m条边，每个点有一个点权，现在要把这n个点拆开，拆下一个节点要消耗的能量是现在剩下的点中与要拆节点相连的所有节点的权值之和。问把这个图拆成单独的n个节点的最小消耗

思路：要把图拆开一共要拆下n-1个节点，每次拆除当前剩余节点中权值最大的点，能保证最后消耗的能量最少

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
#define ll long long
const int N=2005;
int first[N],vis[N],w[N];
int n,m,tot;
struct nod
{
    int id,w;
}a[N];
struct node
{
    int v,nex;
}e[N<<1];
void init()
{
    tot=0;
    memset(first,-1,sizeof(first));
    memset(vis,0,sizeof(vis));
}
void adde(int u,int v)
{
    e[tot].v=v;
    e[tot].nex=first[u];
    first[u]=tot++;
}
bool cmp(nod x,nod y)
{
    return x.w>y.w;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=0;i<n;i++)
        {
            a[i].id=i+1;
            scanf("%d",&a[i].w);
            w[i+1]=a[i].w;
        }
        int u,v;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            adde(u,v);
            adde(v,u);
        }
        sort(a,a+n,cmp);
        int ans=0;
        for(int i=0;i<n-1;i++)
        {
            u=a[i].id;
            vis[u]=1;
            for(int j=first[u];~j;j=e[j].nex)
            {
                v=e[j].v;
                if(!vis[v])  //如果这个点还没被拆下
                    ans+=w[v];
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}