[Codeforces 437C] The Child and Toy

[题目链接]

          https://codeforces.com/contest/437/problem/C

[算法]

         按照点的权值从小到大删点为最优策略

         每条边对答案的贡献为两个端点权值的最小值

         时间复杂度 ： O(N + M)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010

int a[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
int main()
{
        
        int n , m , ans = 0;
        read(n); read(m);
        for (int i = 1; i <= n; i++) read(a[i]);
        for (int i = 1; i <= m; i++)
        {
                int x , y;
                read(x); read(y);
                ans += min(a[x],a[y]);
        }
        printf("%d
",ans);
        
        return 0;
    
}