codeforces437C

最新推荐文章于 2021-04-26 17:59:26 发布

转载最新推荐文章于 2021-04-26 17:59:26 发布 · 177 阅读

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原文链接：http://www.cnblogs.com/gaojunonly1/p/10590906.html

本文介绍了一个关于玩具拆解的算法问题，目标是最小化移除玩具所有部件所需的总能量。通过分析每个部件移除时的能量消耗，提出了一种高效算法，利用边的贡献最小化总能量。

The Child and Toy

CodeForces - 437C

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 v_i. The child spend v_f₁ + v_f₂ + ... + v_{f_k} energy for removing part i where f₁, f₂, ..., f_k 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, ..., v_n (0 ≤ v_i ≤ 10⁵). Then followed m lines, each line contains two integers x_i and y_i, representing a rope from part x_i to part y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i).

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.

sol：容易发现删去一个点等于删掉所有与这个点相连的边，考虑每条边的贡献，比如边<a,b>，肯定取a,b中花费小的更优

Ps：代码极短

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=1005;
int n,m,Cost[N];
#define Pic Picture
int main()
{
    int i,ans=0;
    R(n); R(m);
    for(i=1;i<=n;i++) R(Cost[i]);
    for(i=1;i<=m;i++) ans+=min(Cost[read()],Cost[read()]);
    Wl(ans);
    return 0;
}
/*
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
*/

View Code

转载于:https://www.cnblogs.com/gaojunonly1/p/10590906.html