标签：sign cti comm ima for aced one and sub
InputEach test case will begin with two integers m and n, followed by n integers S 1, S2, S 3 ... S n.
Process to the end of file.
OutputOutput 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.
本题的大致意思为给定一个数组，求其分成m个不相交子段和最大值的问题。
用一个数组去寻找当前段的最大值，用另一个数组保存寻找过的最大值。
// Asimple #include <iostream> #include <algorithm> #include <cstdio> #include <cstdlib> #include <queue> #include <vector> #include <string> #include <cstring> #include <stack> #define INF 0x3f3f3f3f #define mod 2016 using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 1000000+5; int n, m, T, len, cnt, num, Max; int dp[maxn], mmax[maxn]; int a[maxn]; void input() { while( scanf("%d%d", &m, &n)!=EOF ) { for(int i=1; i<=n; i++) { scanf("%d", &a[i]); dp[i] = 0; mmax[i] = 0; } dp[0] = 0; mmax[0] = 0; for(int i=1; i<=m; i++) { Max = -INF; for(int j=i; j<=n; j++) { dp[j] = max(dp[j-1]+a[j], mmax[j-1]+a[j]); mmax[j-1] = Max; Max = max(Max, dp[j]); } } printf("%d\n", Max); } } int main() { input(); return 0; }
感谢kuangbin大神的解析http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html
标签：sign cti comm ima for aced one and sub
原文地址：http://www.cnblogs.com/Asimple/p/7384174.html
