1087. All Roads Lead to Rome (30)

题目：

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<=N<=200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N-1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format "City1 City2 Cost". Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.

Output Specification:

For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommended. If such a route is still not unique, then we output the one with the maximum average happiness -- it is guaranteed by the judge that such a solution exists and is unique.

Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommended route. Then in the next line, you are supposed to print the route in the format "City1->City2->...->ROM".

Sample Input:

6 7 HZHROM 100PKN 40GDN 55PRS 95BLN 80ROM GDN 1BLN ROM 1HZH PKN 1PRS ROM 2BLN HZH 2PKN GDN 1HZH PRS 1

Sample Output:

3 3 195 97HZH->PRS->ROM

注意：

1、这道题目跟1018类似，但是复杂度没有1018那么大，这里直接Dijkstra就可以找出solution（就跟我1018里面贴上的那个有一个case未ac的代码思路一样），而不需要像1018那样先找出所有满足条件的路径再dfs从中寻找solution。

2、大概思路是Dijkstra搜索的时候如果找到了cost更小的路径则选择更小的，如果cost相等则选择total happiness更大的，如果total happiness也相等则选择average happiness更大的（也就是路径中city个数最小的）。

3、需要小心路径数量pathnum的计算，我刚开始是只用一个int变量来记录，结果case 2一直不过，后来改成vector来累计就可以ac了，一定要注意这一点。

代码：

//1087#include<iostream>#include<vector>#define inf 65535using namespace std;vector<vector<int>>G;//graph matrixvector<int>cost;//cost of path from the start to each cityvector<int>happiness;//happiness of each cityvector<int>path;//the solution pathvector<int>totalHap;//total happiness of each routevector<int>city;//the value of all the citisint name2index[20000]={0};//name to indexint n;//the number of citiesint str2num(char *city){return (city[0]-'A')*676+(city[1]-'A')*26+city[2]-'A';}void findpath(){//based on Dijkstra algorithmvector<int>pre;//previous city of each city's pathvector<int>citynum;//number of city in pathvector<int>visited;//visited flagvector<int>pathnum;//number of pathint dest=name2index[str2num("ROM")];//the destination Rome//--------innitiate--------pre.assign(n,0);citynum.assign(n,1);visited.assign(n,0);pathnum.assign(n,1);cost=G[0];totalHap=happiness;visited[0]=1;//--------find path--------int k=0;//current citywhile(1){int min=inf;for(int i=0;i<n;++i){//find the nearist unvisited cityif(!visited[i] && cost[i]<min){min=cost[i];k=i;}}if(k==dest)break;visited[k]=1;for(int i=0;i<n;++i){if(!visited[i] && cost[k]+G[k][i]<cost[i]){//choose the path with minimum costpre[i]=k;cost[i]=cost[k]+G[k][i];totalHap[i]=totalHap[k]+happiness[i];citynum[i]=citynum[k]+1;pathnum[i]=pathnum[k];}else if(!visited[i] && cost[k]+G[k][i]==cost[i]){//choose the path with the maximum total happiness or average happinesspathnum[i] += pathnum[k];if(totalHap[k]+happiness[i]>totalHap[i] ||(totalHap[k]+happiness[i]==totalHap[i] && citynum[k]+1<citynum[i])){pre[i]=k;totalHap[i]=totalHap[k]+happiness[i];citynum[i]=citynum[k]+1;}}}}//--------print the result-----printf("%d %d %d %d\n",pathnum[dest],cost[dest],totalHap[dest],totalHap[dest]/citynum[dest]);while(dest!=0){path.push_back(dest);dest=pre[dest];}printf("%c%c%c",city[0]/676+'A',(city[0]/26)%26+'A',city[0]%26+'A');for(int i=path.size()-1;i>=0;--i)printf("->%c%c%c",city[path[i]]/676+'A',(city[path[i]]/26)%26+'A',city[path[i]]%26+'A');}int main(){int m;char name[5];scanf("%d%d%s",&n,&m,name);int start=str2num(name);city.push_back(start);happiness.assign(n,0);G.resize(n);G[0].assign(n,inf);G[0][0]=0;for(int i=1;i<n;++i){scanf("%s%d",name,&happiness[i]);int c=str2num(name);city.push_back(c);name2index[c]=i;G[i].assign(n,inf);G[i][i]=0;}for(int i=0;i<m;++i){//build the graphchar name0[5];int a,b,c;scanf("%s%s%d",name,name0,&c);a=name2index[str2num(name)];b=name2index[str2num(name0)];G[a][b]=G[b][a]=c;}findpath();return 0;}