注意到有兩個fii-1,也就是說他爹必在它前一個位置即這棵樹退化成一條鏈直接暴力枚舉所有情況再寫一個check函數(shù)用來檢查子串是否合法。順帶提一嘴檢查方法為用一個棧從頭到腳依次壓入字符當(dāng)出現(xiàn)“”時彈出棧頂元素看是否匹配。#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];//這玩意兒目前還沒用}for(inti1;in;i){ans0;for(intl1;ln;l){for(intrl;ri;r)if(check(l,r)){ans;}}m^(i*ans);}coutm;return0;}但是這個方法只能得20分對我來說足夠了說明超時了我們應(yīng)當(dāng)考慮優(yōu)化算法。因?yàn)闃湟呀?jīng)退化成了鏈?zhǔn)浇Y(jié)構(gòu)我們可以想想用dp。咋個用呢如果當(dāng)前字符是’(直接將序號入棧如果當(dāng)前字符是 ‘)’1.棧為空說明無法匹配dp[i]02.棧不為空彈出匹配的左括號位置 pos匹配一對 ()同時 pos 左側(cè)連續(xù)的合法括號串可以拼接進(jìn)來。#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];//這玩意兒目前還沒用}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;}然而還是只有55分。考慮把第一段和第二段結(jié)合一下滿足鏈?zhǔn)浇Y(jié)構(gòu)時用dp不滿足時用個暴力深搜能多騙一些是一些。#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;}這樣就可以再多15分了。但最后還是得寫滿分代碼不然寫這題解沒意義?？梢园裠p遷移到樹上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;}