codeforces 437C The Child and Toy

最新推荐文章于 2021-08-06 11:27:53 发布

转载最新推荐文章于 2021-08-06 11:27:53 发布 · 93 阅读

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原文链接：http://www.cnblogs.com/yxwkf/p/5165428.html

本文探讨了一个儿童节玩具拆解问题，旨在找到移除所有玩具部分所需的最小总能量。通过理解输入输出规范和样例，我们简化了问题并实现了高效算法。此题考察了图论知识和贪心策略的应用。

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.

Sample test(s)

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

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.

翻译

在儿童节这一天，我们的小朋友收到了来自Delayyy的礼物：一个玩具。

只是。小朋友实在是太淘气了。他迫不及待的要拆掉这个玩具。

这个玩具由n个部分和m条绳子组成。每条绳子连接着两个部分，而且不论什么两个部分至多由一条绳子直接相连。为了拆掉这个玩具，小朋友必须移除它的全部n个部分。每次他仅仅能移除一个部分。而且移除须要能量。

每个部分都有一个能量值v[i]，他移除第i部分所需的能量是v[f[1]]+v[f[2]]+...+v[f[k]]，当中f[1],f[2],...,f[k]是与i直接相连（且还未被移除）的部分的编号。

请帮助他找到移除n个部分所需的最小总能量。
Input
第一行包括两个整数n,m(1<=n<=10000;0<=m<=2000)。第二行包括n个整数v[1],v[2],...,v[n](0<=v[i]<=10^5)。接下来有m个行，每一行有两个整数x[i]和y[i]，表示一条连接x[i]和y[i]的绳子（1<=x[i],y[i]<=n; x[i]不等于y[i]）。
Output

输出所需的最小能量。

题解

有没有感觉看了这么长的题目非常累。事实上代码就这么短……

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,m,w[1005];
long long ans=0;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&w[i]);
    for(int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        ans+=min(w[x],w[y]);
    }
    printf("%lld",ans);
    return 0;
}

转载于:https://www.cnblogs.com/yxwkf/p/5165428.html