POJ 1946 Cow Cycling

定義dp[i][j][k]表示第i頭牛領頭，此時消耗j點能量，走了k圈的最少時間

如果此時更換領頭的牛，dp[i][k][k]=min{dp[i-1][j][k],dp[i][k][k]}

否則，dp[i][j][k]=min{dp[i][j][k],dp[i][j-sqr(x)][k-x]+1}

狀態定義完後憑空想轉移不好想，對著題目給的策略自己手推了下第一頭牛和第二頭牛的式子，馬上就能寫出方程了:)

最後枚舉下答案dp[i][j][d]就行

代碼：

#include<iostream>#include<memory.h>#include<string>#include<cstdio>#include<algorithm>#include<math.h>#include<stack>#include<queue>#include<vector>#include<map>#include<ctime>using namespace std;const int inf=1<<30;int dp[21][101][101];int sqr[11]={1,4,9,16,25,36,49,64,81,100};int main(){int i,j,k,l,n,e,d;while(scanf("%d%d%d",&n,&e,&d)!=EOF){for(i=0;i<=n;i++){for(j=0;j<=e;j++){for(k=0;k<=d;k++)dp[i][j][k]=inf;}}dp[0][0][0]=0;for(i=1;i<=n;i++){for(j=0;j<=e;j++){for(k=0;k<=d;k++){if(dp[i-1][j][k]!=inf)dp[i][k][k]=min(dp[i][k][k],dp[i-1][j][k]);}}for(j=1;j<=e;j++)for(k=1;k<=d;k++){ for(l=0;l<10;l++){if(j>=sqr[l]&&k>=l+1){dp[i][j][k]=min(dp[i][j][k],dp[i][j-sqr[l]][k-l-1]+1);}elsebreak;}}}int ans=inf;for(i=1;i<=n;i++){for(j=1;j<=e;j++){if(dp[i][j][d]<ans)ans=dp[i][j][d];}}printf("%d\n",ans);}return 0;}