CODEVS 1048石子归并

#include<stdio.h>using namespace std;int n,w[110],f[110][110];int mmin(int a,int b){    return a<b?a:b;}int summ(int a,int b){    int ssum=0;    for(int k=a;k<=b;k++) ssum+=w[k];    return ssum;}int d(int i,int j){    if(f[i][j]>0) return f[i][j];    else if(i==j) return f[i][j]=0;    else{        int minn=1000000;        for(int k=i;k<j;k++){//注意k<j而不是k<=j!            int o=d(i,k)+d(k+1,j)+summ(i,j);//注意第二段从k+1开始            minn=o<minn?o:minn;        }        return f[i][j]=minn;    }}int main(){    freopen("1048.in","r",stdin);    freopen("1048.out","w",stdout);    cin>>n;    for(int i=1;i<=n;i++) cin>>w[i];    cout<<d(1,n);    return 0;}

题解：记忆化搜索，状态转移方程 d(i,j)=min(d(i,k)+d(k+1,j)+summ(i,j))注意第二段从k+1开始,边界d(i,j)={0|i==j}.