hdu6323(單調決策性DP)
題意:給定一個n*m的矩陣,可以在矩陣內將A個元素改成0,現在要在矩陣內選出B個不相交的寬為m的矩陣,使得這3個矩陣的和最大
這個可以將所有行壓成一個點然後變成序列上的問題,修改可以預處理,顯然對每行來說,應優先修改最小的元素,那麼就可以預處理出a[i][j],代表第i行修改j次後的權值。顯然修改的效果會越來越小,所以f(j)=a[i][j]是一個上凸函式。。
然後設d[p][i][j]為到第i行,選了p個矩形(第p個矩形以第i行結尾),修改了j次的最大值
那麼d[p][i][j]=max{max{d[p][i-1][k]+a[i][j-k]},max{d[p-1][v][k]+a[i][j-k]}}
=max{max(d[p][i-1][k],d[p-1][v][k])+a[i][j-k]}
那麼設g[p][i][k]=max(d[p][i][k],d[p-1][v][k])
d[p][i][j]=max{g[p][i][k]+a[i][j-k]}
複雜度是O(TnABm),顯然會T
而由於a的性質比較特殊,所以這個決策具有單調性
對於2個決策點k<v<j,如果g[p][i][k]+a[i][j-k]<g[p][i][v]+a[i][j-v],由於a[i][x]為上凸函式,所以隨著j增加,v一直比k優,因此可以用單調佇列維護決策,用二分找到決策區間就可以了。。
/* * ┏┓ ┏┓ * ┏┛┗━━━━━━━┛┗━━━┓ * ┃ ┃ * ┃ ━ ┃ * ┃ > < ┃ * ┃ ┃ * ┃... ⌒ ... ┃ * ┃ ┃ * ┗━┓ ┏━┛ * ┃ ┃ Code is far away from bug with the animal protecting * ┃ ┃ 神獸保佑,程式碼無bug * ┃ ┃ * ┃ ┃ * ┃ ┃ * ┃ ┃ * ┃ ┗━━━┓ * ┃ ┣┓ * ┃ ┏┛ * ┗┓┓┏━┳┓┏┛ * ┃┫┫ ┃┫┫ * ┗┻┛ ┗┻┛ */ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<set> #include<bitset> #define inc(i,l,r) for(int i=l;i<=r;i++) #define dec(i,l,r) for(int i=l;i>=r;i--) #define link(x) for(edge *j=h[x];j;j=j->next) #define mem(a) memset(a,0,sizeof(a)) #define ll long long #define eps 1e-8 #define succ(x) (1LL<<(x)) #define lowbit(x) (x&(-x)) #define sqr(x) ((x)*(x)) #define mid (x+y>>1) #define NM 105 #define nm 10005 #define pi 3.1415926535897931 using namespace std; const ll inf=1e9+7; ll read(){ ll x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); return f*x; } int q[nm],qh,qt,b[NM],n,tot,m,_p,v[nm],f[nm]; ll a[NM][3005],g[NM][nm],d[NM][nm],ans; int main(){ //freopen("data.in","r",stdin); int _=read();while(_--){ n=read();tot=read();m=read();_p=read(); inc(i,1,n){ a[i][0]=0; inc(j,1,tot)b[j]=read(),a[i][0]+=b[j]; sort(b+1,b+tot+1); inc(j,1,tot)a[i][j]=a[i][j-1]+max(0,-b[j]); } //inc(i,1,n){inc(j,0,tot)printf("%lld ",a[i][j]);putchar('\n');} mem(g);ans=0; inc(i,1,n)inc(j,0,m)d[i][j]=-inf; inc(p,1,_p){ inc(i,1,n){ q[qh=qt=1]=0;mem(f);mem(v); d[i][0]=max(d[i][0],g[i-1][0]+a[i][0]); g[i][0]=max(g[i][0],d[i][0]); inc(j,1,m){ f[j]=max(f[j],f[j-1]); while(qh<=qt&&f[j]!=q[qh])qh++; int s=j; while(qh<=qt){ s=q[qt]+tot+1; for(int x=j,y=min(m,q[qt]+tot);x<=y;) if(g[i-1][q[qt]]+a[i][mid-q[qt]]<=g[i-1][j]+a[i][mid-j])s=mid,y=mid-1;else x=mid+1; if(s<=v[q[qt]])qt--;else break; } if(s<=m)v[j]=s,f[s]=j,q[++qt]=j; while(qh<=qt&&f[j]!=q[qh])qh++; //printf("%d %d:%d\n",i,j,q[qh]); d[i][j]=max(d[i][j],g[i-1][q[qh]]+a[i][j-q[qh]]); g[i][j]=max(g[i][j],d[i][j]); } } //inc(i,1,n){inc(j,0,m)printf("%lld ",d[i][j]);putchar('\n');}putchar('\n'); inc(i,1,n)inc(j,0,m)g[i][j]=max(g[i][j],g[i-1][j]); } inc(j,0,m)ans=max(ans,g[n][j]); printf("%lld\n",ans); } return 0; }
Problem E. Find The Submatrix
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 43 Accepted Submission(s): 15
Problem Description
Little Q is searching for the submatrix with maximum sum in a matrix of n rows and m columns. The standard algorithm is too hard for him to understand, so he (and you) only considers those submatrixes with exactly m columns.
It is much easier now. But Little Q always thinks the answer is too small. So he decides to reset no more than A cells' value to 0, and choose no more than B disjoint submatrixes to achieve the maximum sum. Two submatrix are considered disjoint only if they do not share any common cell.
Please write a program to help Little Q find the maximum sum. Note that he can choose nothing so the answer is always non-negative.
Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there are 4 integers n,m,A,B(1≤n≤100,1≤m≤3000,0≤A≤10000,1≤B≤3).
Each of the following n lines contains m integers, the j-th number on the i-th of these lines is wi,j(|wi,j|≤109), denoting the value of each cell.
Output
For each test case, print a single line containing an integer, denoting the maximum sum.
Sample Input
2 5 1 0 1 3 -1 5 -1 -2 5 1 1 1 3 -1 5 -1 -2
Sample Output
7 8
Source
2018 Multi-University Training Contest 3
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