SDKD 2017 Summer Team Training #12, tm ( Yet Another Median Task )
阿新 • • 發佈:2017-08-23
cos 位數 const mem bsp sizeof turn sca ret
題目大意:給定一個矩陣,給你多次查詢,查詢一個小矩陣中的數的中位數;
解題思路:完全沒往二分上想,我們可以二分枚舉中位數數值,搜索矩陣中數值小於中位數的個數。
#include <cstdio> #include <iostream> #include <string.h> #include <string> #include <map> #include <queue> #include <vector> #include <set> #include <algorithm> #include<math.h> #include <cmath> #include <stack> #define mem0(a) memset(a,0,sizeof(a)) #define meminf(a) memset(a,0x3f,sizeof(a)) using namespace std; typedef long long ll; typedef long double ld; const int maxn=805,inf=0x3f3f3f3f; const ll llinf=0x3f3f3f3f3f3f3f3f; const ld pi=acos(-1.0L);int a[maxn][maxn]; int p[maxn*maxn]; int main() { int n,i,j,k,l,r,u,d,q; ll m=-llinf,w=llinf; scanf("%d%d",&n,&q); for (i=1;i<=n;i++) { for (j=1;j<=n;j++) { scanf("%d",&a[i][j]); m=max(m,(ll)a[i][j]); w=min(w,(ll)a[i][j]); } }for (i=1;i<=q;i++) { scanf("%d%d%d%d",&u,&l,&d,&r); int lc=w,rc=m,mid,tot=(r-l+1)*(d-u+1),ans; if (tot%2) tot/=2; else tot=tot/2-1; while (lc<=rc) { int cnt=0; mid=(lc+rc)/2; for (j=u;j<=d;j++) { for (k=l;k<=r;k++) { if (a[j][k]<mid) cnt++; } } if (cnt<=tot) lc=mid+1,ans=mid; else { rc=mid-1; } } printf("%d\n",ans); } return 0; }
SDKD 2017 Summer Team Training #12, tm ( Yet Another Median Task )