task1 T2 game
2017-10-02 08:46
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把边权分为两半,加在端点上,那么若一个玩家选取了两个端点,他就能获得该边的边权,若两个玩家各选取了一个端点,那么相减之后相当于没有玩家获得该边的边权。因此把边权分为两半,加在端点上,然后按端点权值排序,交错取即可。
#include<bits/stdc++.h>
#define MAXN 10005
using namespace std;
template <typename T> void read(T &x){
x=0;int f=1;char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
x*=f;
}
int n,m;
double w[MAXN];
inline bool cmp(double x,double y){return x>y;}
int main(){
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
read(n),read(m);
for(int i=1;i<=n;++i) cin>>w[i];
for(int i=1;i<=m;++i){
int x,y;
double v;
read(x),read(y),cin>>v;
v/=2;
w[x]+=v;
w[y]+=v;
}
//for(int i=1;i<=n;++i) cout<<w[i]<<" ";cout<<endl;
sort(w+1,w+n+1,cmp);
//for(int i=1;i<=n;++i) cout<<w[i]<<" ";cout<<endl;
double tot=0.0;
for(int i=1;i<=n;i+=2){
tot+=w[i];
}
for(int i=2;i<=n;i+=2){
tot-=w[i];
}
cout<<tot<<endl;
return 0;
}
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