HDU 4647 Another Graph Game(贪心、边权映射到点权)

题意:

$N,M\le 10^5,N个点M条边,点权W_i、边权C_i \le 10^9 $
$2个人玩游戏轮流选点得到权值,选过的不能再选$
$规定如果一条边的2个端点都被同一个人选到,那么它获得边权$
$假设2个人采取最优策略,输出先手得分-后手得分$

分析:

$考虑没有边的情况,那么显然2个人从大到小选$
$有边,那么就把边权均分到2个点上,如果同1个人选到刚好就加起来了,不同人就抵消了$
$符合题意,避免小数,把权加倍即可$
$时间复杂度O(nlogn)$

代码:

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//
// Created by TaoSama on 2016-03-11
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m;
long long a[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &m) == 2) {
for(int i = 1; i <= n; ++i) scanf("%I64d", a + i), a[i] <<= 1;
for(int i = 1; i <= m; ++i) {
int u, v, w; scanf("%d%d%d", &u, &v, &w);
a[u] += w; a[v] += w;
}
sort(a + 1, a + 1 + n);
long long ans = 0;
for(int i = n; i; --i)
if(i & 1) ans -= a[i];
else ans += a[i];
printf("%I64d\n", ans >> 1);
}
return 0;
}


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