1057-collecting Gold (State compression DP)

Title: Give you a Matrix, ' X ' is your starting position, ' G ' is the location of the treasure, ask at least how many steps can take all the treasure, and finally return to the starting position. Note: When there is no treasure, output 0 ====================================================================================

#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>using namespaceStd;typedefLong LongLL;Const intINF = 1e9+7;Const intMAXN =40000;intdp[maxn][ -], M, N, K;///status, where it is currently locatedCharmaps[ -][ -];structpoint{intx, y;} p[ -], Star;intGetlen (Point A, point B) {intlen = min (abs (a.x-b.x), ABS (a.y-b.y)); Len+ = MAX (ABS (a.x-b.x), ABS (A.Y-B.Y))-Len; returnLen;}intDFS (intStaintX///status, where it is now{    if(dp[sta][x]! =-1)returnDp[sta][x]; DP[STA][X]=INF;  for(intI=0; i<k; i++)    {        if((sta& (1<<i)) && I! =x) {dp[sta][x]= Min (dp[sta][x], DFS (sta-(1&LT;&LT;X), I) +Getlen (P[x],p[i])); }    }   //printf ("dp[%d][%d] =%d\n", STA, X, dp[sta][x]);    returndp[sta][x];}intMain () {intT, CAS =1; scanf ("%d", &T);  while(T--) {scanf ("%d%d", &n, &m); K=0; Memset (DP,-1,sizeof(DP));  for(intI=0; i<n; i++) {scanf ("%s", Maps[i]);  for(intj=0; j<m; J + +)            {                if(Maps[i][j] = ='g') p[k].x= i, p[k++].y =J; if(Maps[i][j] = ='x') star.x= i, star.y =J; }        }         for(intI=0; i<k; i++)        {            intSTA =1<<i; Dp[sta][i]=Getlen (Star, p[i]); }        intAns = INF, Lim = (1&LT;&LT;K)-1;  for(intI=0; i<k; i++)///last place to stayans = min (Ans,dfs (lim,i) +Getlen (Star,p[i])); if(k = =0) ans=0; printf ("Case %d:%d\n", CAS + +, ans); }    return 0;}

1057-collecting Gold (State compression DP)