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=;
 const int INF=0x3f3f3f3f;
 const int MOD= ,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e30;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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=;i<=n;i++){
             scanf("%lf",&v[i]);
         }
         for(i=;i<m;i++){
             scanf("%d%d%lf",&a,&b,&c);
             v[a]+=c/;
             v[b]+=c/;
         }
         sort(v+,v+n+);
         ans=;
         for(i=n;i>=;i-=){
             ans+=v[i]-v[i-];
         }

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