A certain string-processing language allows the programmer to break a string into two pieces. Since this involves copying the old string, it costs n units of time to break a string of n characters into two pieces. Suppose a programmer wants to break a string into many pieces. The order in which the breaks are made can affect the total amount of time used. For example, suppose we wish to break a 20 character string after characters 3, 8, and 10 (numbering the characters in ascending order from the left-hand end, starting from 1). If the breaks are made in left-to-right order, then the first break cost 20 units of time, the second break costs 17 units of time, and the third breaks costs 12 units of time, a total of 49 units of time (see the sample below). If the breaks are made in right-to-left order, then the first break costs 20 units of time, the second break costs 10 units of time, and the third break costs 8 units of time, a total of 38 units of time.
The cost of making the breaks in left-to-right order:
thisisastringofchars (original)thi sisastringofchars (cost:20 units)thi sisas tringofchars (cost:17 units)thi sisas tr ingofchars (cost:12 units) Total: 49 units.
The cost of making the breaks in right-to-left order:
thisisastringofchars (original)thisisastr ingofchars (cost:20 units)thisisas tr ingofchars (cost:10 units)thi sisas tr ingofchars (cost: 8 units) Total: 38 units.
Input:
There are several test cases! In each test case, the first line contains 2 integers N (2<=N<=10000000) and M (1<=M<=1000, M<N). N is the original length of the string, and M is the number of the breaks. The following lines contain M integers Mi (1<=Mi<N) in ascending order that represent the breaking positions from the string's left-hand end.
Output:
For each test case, output in one line the least cost to make all the breakings.
Sample Input:
20 33 8 10
Sample Output:
37
Author: Wei, Qizheng
Source: ZOJ Monthly, June 200和那个石子合并成集合竟然是一模一样，字符串就是把整个分成一个个，也可以看成一个个小石子合成一个大的集合！这其实是一样的！所以，要做的就是把整个分成一个个小的块，再合并起来就得到了答案，核心是一样的！
#include#include #include using namespace std;#define MAXN 1050#define inf 1000000000int prime[MAXN];int dp[MAXN][MAXN],kk[MAXN][MAXN],sum[MAXN],p[MAXN];int main(){ int n,i,k,j,len,m; while(scanf("%d%d",&m,&n)!=EOF) { p[0]=0; for(i=1;i<=n;i++) { scanf("%d",&p[i]); prime[i]=p[i]-p[i-1]; sum[i]=sum[i-1]+prime[i]; } sum[0]=0; n++; prime[n]=m-p[n-1]; sum[n]=sum[n-1]+prime[n]; for(i=0;i<=n;i++) for(j=0;j<=n;j++) { dp[i][j]=inf; } for(i=1;i<=n;i++) { dp[i][i]=0; kk[i][i]=i; } for(len=2;len<=n;len++) { for(i=1;i<=n-len+1;i++) { j=i+len-1; for(k=kk[i][j-1];k<=kk[i+1][j];k++)//此时的k取法不同 { int temp=dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]; if(temp