题目如下：
C. The Child and Toy
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 test(s)
input
4 310 20 30 401 41 22 3
output
40
input
4 4100 100 100 1001 22 32 43 4
output
400
input
7 1040 10 20 10 20 80 401 54 74 55 25 76 41 61 34 31 4
output
160
Note
One of the optimal sequence of actions in the first sample is:
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.
题意：题目大意可以转化为，有许多点和许多线，其中每个点都有一个value（后面说明），每条线连接连接两个点，这样就形成了一张图（可能包含多个子图）。现在要求去掉所有边，求去掉所有边的cost之和的最小值。
思路分析：刚一看到这个题的时候，会很自然的想这些边应该会按照一定顺序去掉（例如，贪心），我们只要找到这个顺序，然后把所有cost加起来就是结果。但是很快发现，这个顺序不是那么容易找到。再仔细一看题目，要求的是最小值，并没有要求得到顺序。那么，是不是存在一种方法直接找到最小值而不用求去掉边的顺序呢？答案是肯定的。仔细分析一下不难发现，当求得最小值的时候，必然是所有的边都已经去掉了。也就是说，无论以什么样的顺序去掉这些边，得到最终的结果时所有边都已经去掉了，而我们就是只要结果。去每一条边的时候，都会有一个代价值cost，必然选所连接的两个点中value最小的那个(假设一条边连接的两个点为A和B，去掉边的时候，选value(A)和value(B)中较小的作为代价值)，把所有的边都去掉，所有cost加起来就是最后的答案。
有了上面的思路，代码就很简洁了，如下：
#includeusing namespace std; int main(){ int n, m, v[1024], res = 0, x, y ; cin >> n >> m ; for(int i = 1; i > v[i] ; while(m--) { cin >> x >> y ; res += min(v[x], v[y]) ; } cout
这个题目告诉我们，看似复杂的问题，背后往往都有一个简单的规律。