UVa 10003 Cutting Sticks (區間dp)

標籤：dp

連結：UVa 10003

題意:給出一根木棍的長度，及木棍上的n個點，要在這n個點處切斷木棍，在切斷木棍時木棍有多長就花費多少代價，求將給定的所有點都切斷的最小代價

分析：這個是區間dp的題，用dp[i][j]數組表示在區間[i,j]內切割木棍的最小代價，

則狀態轉移方程為dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j]+a[j]-a[i]) (i<=k< j)

參考代碼：

#include<stdio.h>#include<string.h>#define M 100000int min(int a,int b){    return a<b?a:b;}int main(){    int a[55],n,i,j,l,r,k,m,dp[55][55];    while(scanf("%d",&m)!=EOF){        if(m==0)            break;        scanf("%d",&n);        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        a[0]=0;              //添加頭尾點        a[n+1]=m;        for(i=0;i<=n;i++){            dp[i][i+1]=0;            for(j=i+2;j<=n+1;j++)                dp[i][j]=M;        }        for(i=1;i<=n+1;i++)               //分段            for(l=0;l<=n+1-i;l++){        //r-l=i<=n+1 -->  l<=n+1-i                r=l+i;                for(k=l+1;k<r;k++)                    dp[l][r]=min(dp[l][r],dp[l][k]+dp[k][r]+a[r]-a[l]);            }        printf("The minimum cutting is %d.\n",dp[0][n+1]);    }    return 0;}

#include<stdio.h>#include<string.h>#define M 100000int a[55],dp[55][55];int min(int a,int b){    return a<b?a:b;}int my_dp(int l,int r){    int i;    if(dp[l][r]!=M)        return dp[l][r];    for(i=l+1;i<r;i++)        dp[l][r]=min(dp[l][r],my_dp(l,i)+my_dp(i,r)+a[r]-a[l]);    return dp[l][r];}int main(){    int i,j,n,m;    while(scanf("%d",&m)!=EOF){        if(m==0)            break;        scanf("%d",&n);        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        a[0]=0;        a[n+1]=m;        for(i=0;i<=n;i++){            dp[i][i+1]=0;            for(j=i+2;j<=n+1;j++)                dp[i][j]=M;        }        my_dp(0,n+1);        printf("The minimum cutting is %d.\n",dp[0][n+1]);    }    return 0;}