题目链接
感觉这题好厉害...根本没想到怎么做。
我们可以把一条边的权值平均分给它的两个端点, 这样, 如果一个人选了两个端点, 那么相加之和等于这条边的权值。 如果是两个人选了, 那么就相互抵消了。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
double a[100005];
int main()
{
int n, m, u, v;
double w;
while(cin>>n>>m) {
for(int i = 1; i<=n; i++)
scanf("%lf", &a[i]);
while(m--) {
scanf("%d%d%lf", &u, &v, &w);
a[u] += w/2;
a[v] += w/2;
}
sort(a+1, a+n+1);
double ans1 = 0, ans2 = 0;
for(int i = 1; i<=n; i++) {
if(i&1)
ans1 += a[i];
else
ans2 += a[i];
}
printf("%.0f
", ans2-ans1);
}
return 0;
}