題意：

給出一些句子，代表一些名詞和動詞的關系，然后有一些查詢，看查詢語句中的關系是否正確。

思路：

按照題意建邊，查詢時找兩個點有沒有路徑即可。

代碼：

#include <cstdio> #include <sstream> #include <iostream> #include <cstring> #include <algorithm> #include <vector> #include <map> using namespace std;map<string,int> mm; map<string,int> nn; int tot; struct E {int to, next; }e[400000]; int head[1000]; int to;void init() {tot = 1;to = 1;mm.clear();nn.clear();memset(head, 0, sizeof(head)); }void add(int u,int v) { // printf("lalalal %d %d\n",u,v);e[to].to = v;e[to].next = head[u];head[u] = to++; }int vis[1000]; int dfs(int u,int v) {vis[u] = 1;if(u == v) return true;for(int i = head[u]; i; i = e[i].next) {int vv = e[i].to;if(!vis[vv]) {if(dfs(vv,v)) {return 1;}}}return 0; }int main() {int t;scanf("%d", &t);getchar();int kase = 1;while (t--) {init();string rans="";printf("Case #%d:\n", kase++);string sentence;while(getline(cin, sentence)){if (sentence == "") {continue;} int senlen = sentence.length();if (sentence[senlen - 1] == '!') {break;} else if (sentence[senlen - 1] == '.') {vector<string> word;stringstream ss(sentence);string nowstring;while (ss >> nowstring) {word.push_back(nowstring);} int wordslen = word.size();word[wordslen - 1] = word[wordslen - 1].substr(0, word[wordslen - 1].length() - 1); if (word[1] == "are") {if(mm[word[0]]==0)mm[word[0]]=tot++;if(mm[word[2]]==0)mm[word[2]]=tot++;add(mm[word[0]],mm[word[2]]);} if (word[1] == "can") {if(mm[word[0]]==0)mm[word[0]]=tot++;if(nn[word[2]]==0)nn[word[2]]=tot++;add(mm[word[0]],nn[word[2]]);} if (word[1] == "which") {if (word[4] == "can") {if(nn[word[3]]==0)nn[word[3]]=tot++;if(nn[word[5]]==0)nn[word[5]]=tot++; // cout<<word[3]<<" "<<word[5]<<endl;add(nn[word[3]],nn[word[5]]);}else if (word[4] == "are") {if(nn[word[3]]==0)nn[word[3]]=tot++;if(mm[word[5]]==0)mm[word[5]]=tot++;add(nn[word[3]],mm[word[5]]);}} }else if (sentence[senlen - 1] == '?') {memset(vis,0,sizeof(vis));int ans;vector<string> word;stringstream ss(sentence);string nowstring;while (ss >> nowstring) {word.push_back(nowstring);} int wordslen = word.size();word[wordslen - 1] = word[wordslen - 1].substr(0, word[wordslen - 1].length() - 1);if (word[0] == "are") {if (word.size() == 3) {if(mm[word[1]]==0)mm[word[1]]=tot++;if(mm[word[2]]==0)mm[word[2]]=tot++;int u=mm[word[1]];int v=mm[word[2]];ans=dfs(u,v);// printf("u=%d v=%d\n",u,v);}else {if(nn[word[4]]==0)nn[word[4]]=tot++;if(mm[word[5]]==0)mm[word[5]]=tot++;int u=nn[word[4]];int v=mm[word[5]];ans=dfs(u,v);// printf("u=%d v=%d\n",u,v);}}if (word[0] == "can") {if (word.size() == 6) {if(nn[word[4]]==0)nn[word[4]]=tot++;if(nn[word[5]]==0)nn[word[5]]=tot++;int u=nn[word[4]];int v=nn[word[5]];ans=dfs(u,v);// printf("u=%d v=%d\n",u,v);}else {if(mm[word[1]]==0)mm[word[1]]=tot++;if(nn[word[2]]==0)nn[word[2]]=tot++;int u=mm[word[1]];int v=nn[word[2]];ans=dfs(u,v);// printf("u=%d v=%d\n",u,v);}}if(ans)rans+="Y";else rans+="M";}}cout<<rans<<endl;}return 0; }

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