标签：include span wan const equal using dex amp alt
D - DZY Loves Modification
As we know, DZY loves playing games. One day DZY decided to play with a n?×?mmatrix. To be more precise, he decided to modify the matrix with exactly koperations.
Each modification is one of the following:
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.
Input
The first line contains four space-separated integers n,?m,?k and p (1?≤?n,?m?≤?103; 1?≤?k?≤?106; 1?≤?p?≤?100).
Then n lines follow. Each of them contains m integers representing aij (1?≤?aij?≤?103) — the elements of the current row of the matrix.
Output
Output a single integer — the maximum possible total pleasure value DZY could get.
Example
2 2 2 2
1 3
2 4
11
2 2 5 2
1 3
2 4
11
Note
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
1 1
0 0
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
-3 -3
-2 -2
题目的意思是,给你一个n*m的矩阵,每个位置有一个数.你可以进行k次操作,每次操作可以选择一行(列),然后把ans累加上这一行(列)的数的和,然后把这一行(列)的每一个数都减去一个固定的值.
要求最大化ans.
我们设R[i]为选择i个(次)行的最优解,C[i]为选择i个(次)列的最优解,由于选择的总次数为K,所以
ans=max(ans,R[i]+C[K-i]).然而这是不对的,因为在考虑R和C的时候是相对独立的,不受另一个因素的影响,而实际上,选择i个行,j个列,会产生i*(K-i)个交点,由于交点的存在,所以还要减去i*(K-i)*p(那个固定值).
所以ans=max(ans,R[i]+C[K-1]-i*(K-i)*p)(要注意数据类型)
那么我们现在来关注如何构造出R和C.由于贪心的想法,我们肯定挑目前和最大的行(列),那么我们可以用两个优先队列来维护一下不就好了.
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<queue> 6 using namespace std; 7 const int maxn=1005,maxK=1000005; 8 int sum_R[maxn],sum_C[maxn]; 9 long long R[maxK],C[maxK]; 10 int n,m,K,p,a[maxn][maxn]; 11 long long ans; 12 struct node{ 13 int x,index; 14 bool operator < (const node u) const {return x<u.x;} 15 }; 16 priority_queue <node> Q_R,Q_C; 17 int read(){ 18 int x=0,f=1; char ch=getchar(); 19 while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘) f=-f; ch=getchar();} 20 while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); 21 return x*f; 22 } 23 int main(){ 24 n=read(),m=read(),K=read(),p=read(),ans=0; 25 for (int i=1; i<=n; i++) 26 for (int j=1; j<=m; j++) a[i][j]=read(); 27 for (int i=1; i<=n; i++){ 28 sum_R[i]=0; 29 for (int j=1; j<=m; j++) sum_R[i]+=a[i][j]; 30 } 31 for (int i=1; i<=m; i++){ 32 sum_C[i]=0; 33 for (int j=1; j<=n; j++) sum_C[i]+=a[j][i]; 34 } 35 R[0]=C[0]=0; 36 for (int i=1; i<=n; i++){node P; P.x=sum_R[i],P.index=i; Q_R.push(P);} 37 for (int i=1; i<=K; i++){ 38 node P=Q_R.top(); Q_R.pop(); 39 R[i]=R[i-1]+P.x,P.x-=m*p,Q_R.push(P); 40 } 41 for (int i=1; i<=m; i++){node P; P.x=sum_C[i],P.index=i; Q_C.push(P);} 42 for (int i=1; i<=K; i++){ 43 node P=Q_C.top(); Q_C.pop(); 44 C[i]=C[i-1]+P.x,P.x-=n*p,Q_C.push(P); 45 } 46 ans=-123456789012345; 47 for (int i=0; i<=K; i++) ans=max(ans,R[i]+C[K-i]-(long long)i*(K-i)*p); 48 printf("%lld",ans); 49 return 0; 50 }
[CodeForces - 447D] D - DZY Loves Modification
标签：include span wan const equal using dex amp alt
原文地址：http://www.cnblogs.com/whc200305/p/7215799.html
