10003 - Cutting Sticks

描述：题目难度为 2，将一条长为n的木棍切成m+1块，但是切割顺序不同，寻找状态方程s[x][y]=min(s[x][y],s[x][i]+s[i][y]+s[y]-s[x]);然后就可以敲代码了代码一（递归，耗时较多）：#include <cstdio>#include <cstring>int num[55],s[55][55];int min(int a,int b){    if(a>b) return b;    else return a;}int dp(int x,int y){    if(y-x==1) return 0;    if(s[x][y]>0) return s[x][y];    s[x][y]=1000000;    for(int k=x+1; k<y; k++) s[x][y]=min(s[x][y],dp(x,k)+dp(k,y)+num[y]-num[x]);    return s[x][y];}int main(){   // freopen("a.txt","r",stdin);    int n,m;    while(scanf("%d",&n)!=EOF)    {        if(!n) break;        memset(s,0,sizeof(s));        scanf("%d",&m);        num[m+1]=n;        for(int i=1; i<=m; i++) scanf("%d",&num[i]);        printf("The minimum cutting is %d.\n",dp(0,m+1));    }    return 0;} 代码二（耗时少）：#include <cstdio>#include <cstring>int num[55],s[55][55];int min(int a,int b){    if(a>b) return b;    else return a;}int main(){   // freopen("a.txt","r",stdin);    int n,m;    while(scanf("%d",&n)!=EOF)    {        if(!n) break;        memset(s,0,sizeof(s));        scanf("%d",&m);        num[m+1]=n;        for(int i=1; i<=m; i++) scanf("%d",&num[i]);        int k=2,x=0;        for( ; x+k<=m+1&&k<=m+1; x++)        {            int y=x+k;            if(!s[x][y]) s[x][y]=1000000;            for(int i=x+1; i<=y; i++) s[x][y]=min(s[x][y],s[x][i]+s[i][y]+num[y]-num[x]);            if(y==m+1&&!x) break;            if(x+k==m+1)            {                x=-1;                k++;            }        }        printf("The minimum cutting is %d.\n",s[0][m+1]);    }    return 0;}