HDU5715 XOR Game Two points + dictionary tree +DP

At that time Astar the time to make only 1 questions, after the game (after a long time to fill, lazy really deadly), found that this is the second simple

Analysis:

This problem can be the minimum XOR of each interval and

When the check is checked, the DP is used to judge, Dp[i][j] represents the first I element into J interval, j is the last element of the last interval

If DP[I][J] is true, indicates that each interval is longer than L, XOR, or greater than mid

Otherwise it is false

Return to Dp[n][m] just fine.

Degree of Complexity O (30^2*NM)

Spit Groove: And the different or related topics are always and the dictionary tree greedy, but also together, remember

#include <stdio.h>#include<algorithm>#include<string.h>using namespacestd;Const intN = 1e4+5; typedefLong LongLL;intN,m,l,t,cas,tot;intA[n];structnode{intsum,nex[2];} P[n* ** A];intNewNode () {++tot; P[tot].sum=0;p [tot].nex[0]=p[tot].nex[1]=-1; returntot;}introot[ A];voidAddintPosintx) {  intnow=root[pos],cur; ++p[now].sum;  for(intI= -; i>=0;--i) {    if(x& (1<<i)) cur=1; ElseCur=0; if(p[now].nex[cur]==-1) P[now].nex[cur]=NewNode (); now=P[now].nex[cur]; ++p[now].sum; }}voidDelintPosintx) {  intnow=root[pos],cur; --p[now].sum;  for(intI= -; i>=0;--i) {     if(x& (1<<i)) cur=1; ElseCur=0; now=P[now].nex[cur]; --p[now].sum; }}intQueryintPosintx) {  intnow=root[pos],ret=0, cur; if(p[now].sum==0)return 0;  for(intI= -; i>=0;--i) {     if(x& (1<<i)) cur=1; ElseCur=0; if(p[now].nex[cur^1]!=-1&&p[p[now].nex[cur^1]].sum!=0) {ret+=(1<<i); now=p[now].nex[cur^1]; }     Elsenow=P[now].nex[cur]; }  returnret;}BOOLdp[n][ A];BOOLCheckintmid) {Tot=0;  for(intI=0; i<m;++i) root[i]=NewNode ();  for(intI=0; i<=n;++i) for(intj=0; j<=m;++j) dp[i][j]=false; dp[0][0]=true; Add (0,0);  for(intI=1; i<=n;++i) {    if(i-l-1>=0){       for(intj=0; j<=m;++j)if(dp[i-l-1][J]) del (j,a[i-l-1]); }     for(intj=1; j<=m;++j) {      intTmp=query (J-1, A[i]); if(tmp>=mid) Add (J,a[i]), dp[i][j]=true; }     }  returndp[n][m];}intMain () {scanf ("%d",&T);  while(t--) {scanf ("%d%d%d",&n,&m,&l);  for(intI=1; i<=n;++i) scanf ("%d", &a[i]), a[i]^=a[i-1]; intL=0, r=1e9+7, ret;  while(l<=R) {      intMid= (l+r) >>1; if(Check (mid)) l=mid+1, ret=mid; Elser=mid-1; } printf ("Case #%d:\n%d\n",++Cas,ret); }  return 0;}

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HDU5715 XOR Game Two points + dictionary tree +DP