注意到有两个fii-1,也就是说他爹必在它前一个位置即这棵树退化成一条链直接暴力枚举所有情况再写一个check函数用来检查子串是否合法。顺带提一嘴检查方法为用一个栈从头到脚依次压入字符当出现“”时弹出栈顶元素看是否匹配。#includebits/stdc.husingnamespacestd;intn,f[500005],ans0;intm;string s;boolcheck(inti,intj){stackcharst;for(intli;lj;l){if(s[l]()st.push(();else{if(st.empty())returnfalse;st.pop();}}returnst.empty();}intmain(){cinn;cins;s s;for(inti1;in;i){cinf[i];//这玩意儿目前还没用}for(inti1;in;i){ans0;for(intl1;ln;l){for(intrl;ri;r)if(check(l,r)){ans;}}m^(i*ans);}coutm;return0;}但是这个方法只能得20分对我来说足够了说明超时了我们应当考虑优化算法。因为树已经退化成了链式结构我们可以想想用dp。咋个用呢如果当前字符是’(直接将序号入栈如果当前字符是 ‘)’1.栈为空说明无法匹配dp[i]02.栈不为空弹出匹配的左括号位置 pos匹配一对 ()同时 pos 左侧连续的合法括号串可以拼接进来。#includebits/stdc.husingnamespacestd;longlongn,f[500005],dp[500005],pos,ans,sum;longlongm;string s;intmain(){cinn;cins;s s;for(longlongi1;in;i){cinf[i];//这玩意儿目前还没用}stacklonglongst;for(longlongi1;in;i){if(s[i](){st.push(i);}else{if(!st.empty()){posst.top();st.pop();dp[i]dp[pos-1]1;}}}for(longlongi1;in;i){sumdp[i];ans^(sum*i);}coutans;return0;}然而还是只有55分。考虑把第一段和第二段结合一下满足链式结构时用dp不满足时用个暴力深搜能多骗一些是一些。#includebits/stdc.husingnamespacestd;intn,m;string s;vectorintG[100005];charval[100005];boolcheck(string t){stackintst;for(intl0;lt.size();l){if(t[l]()st.push(();else{if(st.empty())returnfalse;st.pop();}}returnst.empty();}longlongxdp(){vectorlonglongdp(n1,0);vectorintst;longlongsum0,ans0;for(inti1;in;i){if(s[i-1](){st.push_back(i);dp[i]0;}else{if(!st.empty()){intpostst.back();st.pop_back();dp[i]dp[post-1]1;}else{dp[i]0;}}sumdp[i];ans^(1LL*i*sum);}returnans;}longlongdfs(intu,string path){path.push_back(val[u]);intLpath.size();intk0;for(intl0;lL;l){for(intrl;rL;r){string subpath.substr(l,r-l1);if(check(sub))k;}}longlongans1LL*u*k;for(inti0;iG[u].size();i){intvG[u][i];ans^dfs(v,path);}returnans;}intmain(){cinn;cins;for(inti0;in;i){val[i1]s[i];}boolisftrue;vectorintf(n1);for(inti2;in;i){intx;cinx;f[i]x;if(f[i]!i-1)isffalse;G[f[i]].push_back(i);}longlongans;if(isf){ansxdp();}else{ansdfs(1,);}coutans;return0;}这样就可以再多15分了。但最后还是得写满分代码不然写这题解没意义。可以把dp迁移到树上dp[u]dp[f[m]]1。#includebits/stdc.husingnamespacestd;longlongn,sum0,ans0;string s;vectorlonglongG[500005];longlongf[500005];longlongdp[500005];vectorlonglongst;voiddfs(longlongu){longlongoldsumsum;longlongm-1;if(s[u-1](){st.push_back(u);dp[u]0;}else{if(!st.empty()){mst.back();st.pop_back();dp[u]dp[f[m]]1;}else{dp[u]0;}}sumdp[u];ans^(1LL*u*sum);for(longlongi0;i(longlong)G[u].size();i){dfs(G[u][i]);}sumoldsum;if(s[u-1](){st.pop_back();}else{if(m!-1){st.push_back(m);}}}intmain(){cinn;cins;f[1]0;for(inti2;in;i){cinf[i];G[f[i]].push_back(i);}dfs(1);coutans;return0;}