hdu1024(最大m串子序列)
m的范围没给,很坑爹
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12747 Accepted Submission(s): 4202
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
8
Hint
Huge input, scanf and dynamic programming is recommended.
Author
JGShining(极光炫影)
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <algorithm> #include <math.h> #include <map> #include <queue> #include <sstream> #include <iostream> using namespace std; #define INF 0x3fffffff typedef __int64 LL; #define N 1000100 LL dp[N][2]; LL g[N]; int main() { //freopen("//home//chen//Desktop//ACM//in.text","r",stdin); //freopen("//home//chen//Desktop//ACM//out.text","w",stdout); int m,n; while(scanf("%d%d",&m,&n)!=EOF) { for(int i=1;i<=n;i++) { g[i]=-10000000000000; dp[i][0]=dp[i][1]=-10000000000000; } int a=0,b=1; for(int i=1;i<=n;i++) { swap(a,b); int tmp; scanf("%d",&tmp); for(int j=m;j>=1;j--) { if(j>i) continue; dp[j][a]=max(dp[j][b],g[j-1])+tmp; if(dp[j][a] > g[j]) g[j] = dp[j][a]; } } printf("%I64d\n",g[m]); } return 0; }