HDU 1024 Max Sum Plus Plus[dp](最大m子段和)
Problem Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3 2 6 -1 4 -2 3 -2 3
Sample Output
6 8HintHuge input, scanf and dynamic programming is recommended.
题意:给n个数,要找出m个互不相交的子段,使得他们的和最大。输出最大和。
取某一个数有两种方式:要么并入上一个数所属的段,要么作为一个新段的开头(此时上一个数可不要)。
d[i][j]表示取了第i个数后,前i个数构成j个段的最大和。(默认约束j<=i)
d[i][j] = max { d[i-1][j], d[k][j-1] } + a[i] , 其中,j-1 <= k <= i-1
优化1:方程中除了i和j有循环外,k也有一个循环。这个循环是用来查找到前i-1个数构成j-1个子段的最大和,既然如此我们直接用一个数组记下来就好了。这样还有一个好处,就是我们可以使用滚动数组来进行递推。
优化2:我们最后一步要递推的是d[n][m],这个状态由哪些状态转移过来呢?
只由d[m-1 ... n-1][m-1]和d[n-1][m](如果这些状态合法的话)。那么我们就完全不需要管d[n-1][m-2]是什么东西(我们要求前n个数的m子段和,那么知道前n-1个数的m-2子段和是没用的)。既然d[n-1][m-2 ... 1]是不需要的状态,那么d[n-2][m-3 ... 1]也都是不需要的状态,对j采用逆向递推,就可以剪枝了。
注意这个dp方程是选了第i个数的情况。第n个数可选可不选,前面的也是如此,因此要在递推过程中将最大的答案记录下来。
#include<iostream> #include<cassert> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<string> #include<iterator> #include<cstdlib> #include<vector> #include<stack> #include<queue> #include<map> #include<set> using namespace std; #define debug(x) cout<<"debug "<<x<<endl; #define rep(i,f,t) for(int i = (f),_end_=(t); i <= _end_; ++i) #define rep2(i,f,t) for(int i = (f),_end_=(t); i < _end_; ++i) #define dep(i,f,t) for(int i = (f),_end_=(t); i >= _end_; --i) #define dep2(i,f,t) for(int i = (f),_end_=(t); i > _end_; --i) #define clr(c, x) memset(c, x, sizeof(c) ) typedef long long int64; const int64 INF = 1e15; const double eps = 1e-8; //***************************************************** const int maxn = 1e6+10; int64 *pre,*cur; int64 d[2][maxn]; int64 a[maxn]; int64 mx[maxn]; //优化1 要用的数组 int main() { int n,m; while(scanf("%d%d",&m,&n) == 2){ rep(i,1,n)scanf("%I64d",a+i); if(n==1) { printf("%I64d\n",a[1]); continue; } int64 ans = -INF; pre = d[0]; cur = d[1]; mx[1] = pre[1] = a[1]; mx[0] = pre[0] = 0; rep(i,2,n) { mx[i] = cur[i] = pre[i-1]+a[i]; dep(j,min(m,i-1),1) { if(i-j>n-m)break; //优化2 cur[j] = max(pre[j],mx[j-1])+a[i]; mx[j] = max(mx[j], cur[j]); } if(i>=m) ans = max(ans,cur[m]); swap(cur,pre); } printf("%I64d\n",ans); } return 0; }
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