吉林区域赛 I - Strength
链接：https://vjudge.net/contest/386991#problem/I
题解链接：https://blog.csdn.net/bailichuan266/article/details/96277895
1 #include<bits/stdc++.h> 2 using namespace std; 3 struct node{ 4 int s,t; 5 }a[100005]; 6 int b[100005]; 7 int vis[100005]; 8 int cmp(node x,node y){ 9 return x.s<y.s; 10 } 11 int main(){ 12 int T; 13 scanf("%d",&T); 14 for(int j=1;j<=T;j++){ 15 memset(vis,0,sizeof(vis)); 16 int n,m; 17 scanf("%d%d",&n,&m); 18 for(int i=0;i<n;i++) 19 scanf("%d",&b[i]); 20 for(int i=0;i<m;i++) 21 scanf("%d",&a[i].s); 22 for(int i=0;i<m;i++) 23 scanf("%d",&a[i].t); 24 sort(a,a+m,cmp); 25 sort(b,b+n); 26 long long ans1=0,ans2=0; 27 int l=0,r=0; 28 while(l<n&&r<m){ 29 if(b[l]>=a[r].s){ 30 if(a[r].t==0){ 31 ans1+=b[l]-a[r].s; 32 } 33 vis[l]=1; 34 l++; 35 r++; 36 } 37 else l++; 38 } 39 if(r==m){ 40 for(int i=0;i<n;i++) 41 if(!vis[i]) 42 ans1+=b[i]; 43 } 44 else{ 45 l=0,r=n-1; 46 while(r>=0&&l<m){ 47 if(b[r]>=a[l].s&&a[l].t==0){ 48 ans2+=b[r]-a[l].s; 49 r--; 50 l++; 51 } 52 else l++; 53 } 54 } 55 printf("Case %d: %lld\n",j,max(ans1,ans2)); 56 } 57 }
