Codeforces Round #250 (Div. 2) -C. The Child and Toy

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:

• 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.

思路：要花费最少的能量，那末每次就取最少的能量即可

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=10005;int v[maxn];int n,m;  int main(){int i,j;while(scanf("%d %d",&n,&m)!=EOF){for(i=1;i<=n;i++)scanf("%d",&v[i]);int x,y,sum=0;while(m--){scanf("%d %d",&x,&y);sum+=min(v[x],v[y]);}cout<<sum<<endl;}return 0;}