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题目大意已知一个BFS的遍历顺序每个结点的子节点是按照升序从左到右排列的求树的最少的层数思路将每个升序的子序列接到树的目前的根结点并且该子序列的长度即为下次可以接的子树的个数#pragmaGCCoptimize(-Ofast,-funroll-all-loops)#includecstdio#includealgorithm#includeiostream#includecstring#includemath.husingnamespacestd;#defineendl\n#defineINF0x3f3f3f3f// #define int long long#definedebug(a)cout#aaendl;typedeflonglongll;constdoublePIacos(-1.0);constdoubleeexp(1.0);constintM1e97;constintN2e57;inlineintmymax(intx,inty){returnxy?x:y;}inlineintmymin(intx,inty){returnxy?x:y;}intT,n,a[N];intans,now,cnt,sum;voidsolve(){cinn;anssum1;cntnow0;cina[1];for(inti2;in;i){cina[i];if(a[i]a[i-1])now;else{cnt;if(cntsum){ans;cnt0;sumnow;}}}coutansendl;return;}signedmain(){intT;cinT;while(T--)solve();return0;}