cf437C 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 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.

打cf开黑就是爽……风骚的结构体……

就是每次找个最大的搞掉

#include <vector>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <bitset>
#include <map>
#include <ctime>
#include <cassert>
#include <set>

using namespace std;
typedef pair<int, int> pi;
typedef long long ll;

const int N = 1005;

int cost[N];
int A[N];
bool done[N];

vector< int > adj[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m, a, b;
    cin >> n >> m;
    
    for(int i = 1; i <= n; i++){
        cin >> A[i];
    }
    
    for(int i = 0; i < m; i++){
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    set< pi, greater< pi > > S;
    for(int i = 1; i <= n; i++){
        int& cur = cost[i];
        for(int j = 0; j < adj[i].size(); j++){
            int to = adj[i][j];
            cur += A[to];
        }
        S.insert(pi(A[i], i));
    }
    set< pi, greater< pi > >::iterator it;
    ll tot = 0;
    while(!S.empty()){
        pi cur = *S.begin();
        done[cur.second] = true;
        tot += cost[cur.second];
        S.erase(S.begin());
        for(int i = 0; i < adj[cur.second].size(); i++){
            int to = adj[cur.second][i];
            if(done[to]) continue;
            it = S.find(pi(A[to], to));
            assert(it != S.end());
            S.erase(it);
            cost[to] -= A[cur.second];
            S.insert(pi(A[to], to));
        }
    }
    
    cout << tot << endl;
    
    return 0;
}

　　

其实最简单的代码是这样的：

我当时就跪下了……