HDU-4647 Another Graph Game 贪心,博弈

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4647

　　注意这题两人的决策是想要使得自己的分数与对方的差值最大。。

　　注意到数据范围，显然是贪心之类的，如果没有变那么很简单，如果有边，那么我们进行拆边，把边的权值的一半加到所连的点上。然后排个序贪心。。

//STATUS:C++_AC_218MS_1020KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD= 1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

double v[N];
int n,m;

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,a,b;
    double c,ans;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++){
            scanf("%lf",&v[i]);
        }
        for(i=0;i<m;i++){
            scanf("%d%d%lf",&a,&b,&c);
            v[a]+=c/2;
            v[b]+=c/2;
        }
        sort(v+1,v+n+1);
        ans=0;
        for(i=n;i>=0;i-=2){
            ans+=v[i]-v[i-1];
        }

        printf("%.0f
",ans);
    }
    return 0;
}