POJ-1013-Counterfeit Dollar -暴力水题

http://poj.org/problem?id=1013

这代码写得太挫太长了。。。。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>using namespace std;  struct node{int len;char ip1[105];char ip2[105];char ip3[105]; };node input[5];bool cmp(node a,node b){return a.len<b.len;}char tm[105];char nm[105];char result[105]; int value[15];int vis[15];int vis2[15];set<int>::iterator it;set<int> ans;set<int> ans2;int main(){  int i,j;int t;cin>>t;while(t--){memset(value,0,sizeof(value));memset(vis2,0,sizeof(vis2));memset(vis,0,sizeof(vis));ans.clear();ans2.clear();for (i=1;i<=3;i++){scanf("%s %s %s",input[i].ip1,input[i].ip2,input[i].ip3);input[i].len=strlen(input[i].ip1);} sort(input+1,input+1+3,cmp);  //从长度最短的开始选可保证尽可能得到答案(set里面元素尽可能少)，但是本题说明了必然有答案，所以排序不排序都不影响结果for (i=1;i<=3;i++){int n=input[i].len;strcpy(tm,input[i].ip1);strcpy(nm,input[i].ip2);strcpy(result,input[i].ip3);if (strcmp(result,"even")==0){for (j=0;j<n;j++) vis2[nm[j]-'A'+1]=vis2[tm[j]-'A'+1]=vis[nm[j]-'A'+1]=vis[tm[j]-'A'+1]=1;}elseif (strcmp(result,"up")==0) {//if ans is lightif (ans.size()==0){for (j=0;j<n;j++) ans.insert(nm[j]-'A'+1); //insert all possible value; (must contain the ans)}else{for (it=ans.begin();it!=ans.end();it++){if (std::find(nm,nm+n,(*it+'A'-1))==tm+n ) //cannot findvis[*it]=1;  // mark to delete elsevalue[*it]++;}  }for (j=0;j<n;j++) vis[tm[j]-'A'+1]=1;  //those impossible ans//if ans is heavyif (ans2.size()==0){for (j=0;j<n;j++) ans2.insert(tm[j]-'A'+1); //insert all possible value; (must contain the ans)}else{for (it=ans2.begin();it!=ans2.end();it++){if (std::find(tm,tm+n,(*it+'A'-1))==tm+n )   //cannot findvis2[*it]=1;  // mark to delete elsevalue[*it]++;   }  }for (j=0;j<n;j++) vis2[nm[j]-'A'+1]=1;     //those impossible ans} elseif (strcmp(result,"down")==0) {//if ans is lightif (ans.size()==0){for (j=0;j<n;j++) ans.insert(tm[j]-'A'+1); }else{for (it=ans.begin();it!=ans.end();it++){ if (std::find(tm,tm+n,(*it+'A'-1))==tm+n )  vis[*it]=1;    elsevalue[*it]++;}  }for (j=0;j<n;j++) vis[nm[j]-'A'+1]=1;//if ans is heavyif (ans2.size()==0){for (j=0;j<n;j++) ans2.insert(nm[j]-'A'+1);  }else{for (it=ans2.begin();it!=ans2.end();it++){if (std::find(nm,nm+n,(*it+'A'-1))==tm+n )  vis2[*it]=1;  elsevalue[*it]++;}  }  for (j=0;j<n;j++) vis2[tm[j]-'A'+1]=1;}}for (j=1;j<=12;j++){if (vis[j])ans.erase(j);if (vis2[j])ans2.erase(j);}   int op=-1;if (ans.size()==1&&ans2.size()==1)     // special condition{int t1=*ans.begin();int t2=*ans2.begin();if (value[t1]>value[t2])op=1;elseop=2;if (op!=1)printf("%c is the counterfeit coin and it is heavy.\n",(*ans2.begin()+'A'-1));else printf("%c is the counterfeit coin and it is light.\n",(*ans.begin()+'A'-1)); continue;}if (ans.size()!=1)printf("%c is the counterfeit coin and it is heavy.\n",(*ans2.begin()+'A'-1));else printf("%c is the counterfeit coin and it is light.\n",(*ans.begin()+'A'-1));}return 0;}