codevs3002 石子归并 3

http://blog.csdn.net/noiau/article/details/72514812

#include<cstdio>
#include<cstring>
const int MAXN=1e5+10,INF=1e8+10;
using namespace std;
inline char nc()
{
    static char buf[MAXN],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin)),p1==p2?EOF:*p1++;
}
inline int read()
{
    char c=nc();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}
    return x*f;
}
int dp[3001][3001],sum[MAXN],s[3001][3001];
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    int N=read();
    for(int i=1;i<=N;i++) sum[i]=read(),sum[i]+=sum[i-1],s[i][i]=i;
    for(int i=N;i>=1;i--) 
    {
        for(int j=i+1;j<=N;j++)
        {
            int mn=INF,mnpos=0;
            for(int k=s[i][j-1];k<=s[i+1][j];k++)
            {
                if(dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1] < mn)
                {
                    mn=dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];
                    mnpos=k;
                }
            }
            dp[i][j]=mn;
            s[i][j]=mnpos;
        }
    } 
    printf("%d",dp[1][N]);
    return 0;
}