hdu5375 Gray Code (DP)

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=5375

The main idea: give you a binary string, with '? ' The location can be determined by you to fill in ' 1 ' or ' 0 ', complete after the conversion into gray code, each location has a weight a[i], only gray code for ' 1 ' bit can add weights, ask you the final weights and the maximum number of.

Gray code means Baidu can be a bit, here can be popular point: For such a string, if I position is 1, then the number of his back will change (0 to 1, 1 to 0), note just look at the initial string. For example: The initial string is 110, then the change is 101, not 100.

IDEA: Consider a DP solution, Dp[i][j] indicates the maximum fraction (j take 0 or 1) for the first position of J.

Then there is Dp[i][0]=max (Dp[i-1][0],dp[i-1][1]+a[i]);

Dp[i][1]=max (dp[i-1][0]+a[i],dp[i-1][1]);

When J is? , the J may be 0 or 1, so long both are considered to be processed again.

Note the initialization.

#include <stdio.h> #include <math.h> #include <string.h> #define MAX (b) a>b?a:bint dp[200005][2] , A[200005];int Main () {int T,i,j,l,k,t=0;char s[200005],c[200005];scanf ("%d", &t), while (t--) {t++;scanf ("%s", s); L=strlen (s); for (i=0;i<l;i++) c[i+1]=s[i];for (i=1;i<=l;i++) scanf ('%d ', &a[i]); for (i=0;i<200005;i++) { dp[i][0]=-999999;dp[i][1]=-999999;} if (c[1]== ' 0 ') dp[1][0]=0;if (c[1]== ' 1 ') dp[1][1]=a[1];if (c[1]== '? ') {dp[1][0]=0;dp[1][1]=a[1];} for (i=2;i<=l;i++) {if (c[i]== ' 0 ') {Dp[i][0]=max (dp[i-1][0],dp[i-1][1]+a[i]);} if (c[i]== ' 1 ') {Dp[i][1]=max (dp[i-1][0]+a[i],dp[i-1][1]);} if (c[i]== '? ') {Dp[i][0]=max (dp[i-1][0],dp[i-1][1]+a[i]);DP [I][1]=max (Dp[i-1][0]+a[i],dp[i-1][1]);}} printf ("Case #%d:", T), if (c[l]== ' 0 ') printf ("%d\n", dp[l][0]), else if (c[l]== ' 1 ') printf ("%d\n", dp[l][1]), Else printf ( "%d\n", Max (dp[l][0],dp[l][1]));} return 0;}

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hdu5375 Gray Code (DP)