bzoj2091【Poi2010】The Minima Game
2016-12-11 10:41
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直接dp就好了
每个人肯定会去选最大的,用dp[i]表示选了后i个点时先手-后手的最大值(因为从后往前扫才好转移啊 QwQ~)
dp[i]=max(c[j]-dp[j-1]),(j<=i)
直接维护max值就好了~
每个人肯定会去选最大的,用dp[i]表示选了后i个点时先手-后手的最大值(因为从后往前扫才好转移啊 QwQ~)
dp[i]=max(c[j]-dp[j-1]),(j<=i)
直接维护max值就好了~
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> using namespace std; const int Mx=1000010; int n,c[Mx]; long long maxn,dp[Mx]; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&c[i]); sort(c+1,c+1+n); for(int i=1;i<=n;i++) maxn=max(maxn,c[i]-dp[i-1]),dp[i]=maxn; cout<<dp <<endl; return 0; }
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