POJ 3276 Face the right (common technique-ruler)

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"The main idea": N cattle lined up in a row of 1<=n<=5000. Each cow is either forward or backward. In order for all cows to face the front, the farmer can turn the K-head continuous cow to 1<=k<=n each time, the minimum number of operations m and the corresponding minimum k.

"Idea": because the switching interval reversal order has no effect on the results, so from left to right for the need to flip the cow to reverse, while recording the effect of the other cows in the interval cal sum, for the last part of the interval can not be flipped check whether there is a reverse bull, if there is a scheme failure. The thought of this problem deserves careful consideration, and often there is an infinite state, which becomes a finite state.

Code:

/***************poj 3276 (sq. ft.) *face the right way***************/#include <iostream> #include <stdio.h># Include <string.h>using namespace Std;const int n=5005;int t,n,m;int f[n]; Flip state, 1 flip, 0 not flip int dir[n];    Input status f<>0 b<>1char op[2];int calc (int k) {//enum K, flip order from left to right memset (f,0,sizeof (F));   int i,res=0;    Number of consecutive flips int sum=0;            for (i=0; i+k<=n; ++i) {if ((dir[i]+sum) &1) {//flip odd number of times change state, even several times do not change res++;        F[i]=1;  } Sum+=f[i];    Ruler if (i-k+1>=0) sum-=f[i-k+1];        } for (; i<n; ++i) {//Determine if the remainder has a reverse bull if ((dir[i]+sum) &1) return-1;    else if (i-k+1>=0) sum-=f[i-k+1]; } return res;    void Solve () {int k=1,m=n;        for (int i=1; i<=n; ++i) {int M=calc (i);            if (m>0&&m<m) {m=m;        K=i; }} printf ("%d%d\n", k,m);}     int main () {while (scanf ("%d", &n)!=eof) {for (int i=0; i<n; ++i) {scanf ("%s", op);       if (op[0]== ' F ') dir[i]=0;        else dir[i]=1;    } solve (); } return 0;}

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POJ 3276 Face the right (common technique-ruler)