题目大意:
ai,ai+1,ai+2...
变成
bi,bi+1,bi+2..
不计顺序,增加和减少a数组均有代价。
题解:贪心+排序
小的对应小的
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define N 25009 #define LL long long using namespace std; int n,x,y; LL ans; int a[N],b[N]; int main(){ scanf("%d%d%d",&n,&x,&y); for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]); sort(a+1,a+n+1);sort(b+1,b+n+1); for(int i=1;i<=n;i++){ if(a[i]==b[i])continue; if(a[i]>b[i])ans+=y*(a[i]-b[i]); if(b[i]>a[i])ans+=x*(b[i]-a[i]); } cout<<ans<<endl; return 0; }