Dire Wolf---hdu5115 (interval dp)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5115

Test Instructions: There is a row of wolves, each wolf has a damage a, and a damage B. When you kill a wolf, you will be harmed by the wolf's damage A and the wolf on both sides of the wolf. If a wolf in a position is killed, the wolf on the left will receive B from the right wolf, because the two wolves are adjacent. The minimum cost of killing a row of wolves.

Solution: set DP[I][J] to eliminate the cost of the I to J Wolf, enumeration K as the last killed Wolf, at this time will be affected by a[k] and B[i-1] b[j+1] To minimize the damage the transfer equation can be listed: Dp[i][j]=min (Dp[i][j], dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]) dp[i][i]=a[i]+b[i-1]+b[j+1];

#include <stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<math.h>using namespacestd;#defineN 220#defineINF 0XFFFFFFFintMain () {intT, N, A[n], b[n], dp[n][n], t=1; scanf ("%d", &T);  while(t--) {memset (A,0,sizeof(a)); memset (b,0,sizeof(b)); scanf ("%d", &N);  for(intI=1; i<=n; i++) scanf ("%d", &A[i]);  for(intI=1; i<=n; i++) scanf ("%d", &B[i]);  for(intI=1; i<=n; i++)        {             for(intJ=i; j<=n; J + +) Dp[i][j]=INF; ///dp[i][i]=a[i]+b[i-1]+b[i+1];        }         for(intL=0; l<=n; l++)        {             for(intI=1; i+l<=n; i++)            {                intj=i+l;  for(intK=i; k<=j; k++) {Dp[i][j]=min (Dp[i][j], dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]); }}} printf ("Case #%d:%d\n", t++, dp[1][n]); }    return 0;}

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Dire Wolf---hdu5115 (interval dp)