PAT_1087. All Roads Lead to Rome

#include <iostream>#include <string>#include <string.h>#include <vector>#include <map>#include <algorithm>using namespace std;struct City{string name;int happiness;}city[205];  //数字与城市的映射map<string,int> m;  //城市名字与数字的映射int N,K,idx=1;  int G[205][205];  //邻接矩阵表示图bool vis[205];  //在当前dfs状态下某些点是否被访问过。int minDist=9999999,maxHappiness=0,numCity=0,nextVertex[205],path[205];  //用来记录最优解情况下的各个数据vector<int> diffRoute;//用来记录可以到达终点的每一条路径的长度void dfs(int v,int dist,int happiness,int depth){vis[v]=true;//cout<<"v="<<v<<" city[v].name="<<city[v].name<<",d="<<dist<<",h="<<happiness<<",depth="<<depth<<endl;if(dist>minDist)return;if(v==m["ROM"]){diffRoute.push_back(dist);if(dist<minDist){minDist=dist;maxHappiness=happiness;numCity=depth;memcpy(path,nextVertex,sizeof(path));}else if(dist==minDist&&happiness>maxHappiness){maxHappiness=happiness;numCity=depth;memcpy(path,nextVertex,sizeof(path));}else if(dist==minDist&&happiness==maxHappiness&&happiness/depth>maxHappiness/numCity){numCity=depth;memcpy(path,nextVertex,sizeof(path));}return;}for(int i=1;i<idx;i++){if(vis[i]==true||G[v][i]==-1)continue;nextVertex[v]=i;dfs(i,dist+G[v][i],happiness+city[i].happiness,depth+1);vis[i]=false;}}void outputRoute(int P[]){int tempV=1;while(tempV!=m["ROM"]){cout<<city[tempV].name<<"->";tempV=P[tempV];}cout<<"ROM"<<endl;}int main(){for(int i=0;i<205;i++)for(int j=0;j<205;j++)G[i][j]=-1;string st;cin>>N>>K>>st;m[st]=idx;city[1].name=st;city[1].happiness=0;idx++;string cityName;int hp;for(int i=0;i<N-1;i++){cin>>cityName>>hp;m[cityName]=idx;city[idx].name=cityName;city[idx].happiness=hp;idx++;}string from,to;int dist;for(int i=0;i<K;i++){cin>>from>>to>>dist;G[m[from]][m[to]]=dist;G[m[to]][m[from]]=dist;}dfs(1,0,0,0);int differentRoutes=0;sort(diffRoute.begin(),diffRoute.end());for(int i=0;i<diffRoute.size();i++){if(diffRoute[i]==diffRoute[0])differentRoutes++;elsebreak;}cout<<differentRoutes<<" "<<minDist<<" "<<maxHappiness<<" "<<maxHappiness/numCity<<endl;outputRoute(path);}

这种类型的题做了不少了，如PAT1030 PAT1018 PAT1003 

这题的思路就是先把string类型代表的城市映射成int型，然后解法就与上面几道的思路相同。

采取DFS的策略，在遇到终点（ROM）之后判断是否为更优路径（比较路径，幸福值和，平均幸福和）

这题稍显麻烦的是要求出最短cost的路径总条数，我们只需要把所有可达终点的路径的长度都保存下来，排个序（升序），

那么与第一个路径长度相同的路径都要计入总条数内。

注意保存路径点的方式，我们用两个int 数组 path[] 和 nextVertex[]来保存，nextVertex用于dfs过程中的路径动态更新，path用于每次寻得最新更优解之后

将nextVertex数组中记录的点保存下来，这样dfs完毕之后path中保存的就是最优路径。注意memcpy的用法。