1087. All Roads Lead to Rome (30)

 

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 HZH

ROM 100

PKN 40

GDN 55

PRS 95

BLN 80

ROM GDN 1

BLN ROM 1

HZH PKN 1

PRS ROM 2

BLN HZH 2

PKN GDN 1

HZH PRS 1

Sample Output:

3 3 195 97

HZH->PRS->ROM

  1 #include<stdio.h>

  2 #include<map>

  3 #include<string>

  4 #include<string.h>

  5 #include<stack>

  6 using namespace std;

  7 #define MAX 210

  8 int INF = 1000000;

  9 int HappyVal[210];

 10 int visit[MAX];

 11 int Grap[MAX][MAX];

 12 int d[MAX];

 13 int h[MAX];

 14 int num[MAX];

 15 int pre[MAX];

 16 int Count[MAX];

 17 

 18 void Dijkstra(int Begin,int NodeNum)

 19 {

 20     d[Begin] = 0;

 21     h[Begin] = HappyVal[Begin];

 22     num[Begin] = 1;

 23     Count[Begin] = 0;

 24     for(int i = 0;i < NodeNum ;i++)

 25     {

 26         int index = -1;

 27         int MIN = INF;

 28         for(int j = 0 ;j <NodeNum ;j++)

 29         {

 30             if(!visit[j] && d[j] < MIN)

 31             {

 32                 index = j;

 33                 MIN = d[j];

 34             }

 35         }

 36 

 37         if(index == -1) return ;

 38         visit[index] = true;

 39         for(int v = 0 ;v <NodeNum ;v++)

 40         {

 41             if(!visit[v] && Grap[index][v]!=INF)

 42             {

 43                 if(d[index]+Grap[index][v]<d[v])

 44                 {

 45                     d[v] = d[index]+Grap[index][v];

 46                     num[v] = num[index];

 47                     h[v] = h[index] + HappyVal[v];

 48                     pre[v] = index;

 49                     Count[v] = Count[index] +1;

 50                 }

 51                 else if(d[index]+Grap[index][v]==d[v])

 52                 {

 53                     num[v] = num[v] + num[index];

 54 

 55                     if(h[v] < h[index] + HappyVal[v])

 56                     {

 57                         h[v] = h[index] + HappyVal[v];

 58                         Count[v] = Count[index] +1;

 59                         pre[v] = index;

 60                     }

 61                     else if( h[v] == h[index] + HappyVal[v] && (double)(h[index] + HappyVal[v])/(Count[index]+1) > (double)h[v]/Count[v])

 62                     {

 63                         Count[v] = Count[index] +1;

 64                         pre[v] = index;

 65                     }

 66                 }

 67             }

 68         }

 69     }

 70 

 71 }

 72 

 73 int main()

 74 {

 75     int i,j,N,K,happy,ROM;

 76     char Begin[4],tem[4];

 77     scanf("%d%d%s",&N,&K,Begin);

 78     map<string,int> mm;

 79     map<int,string> mm2;

 80     mm[Begin] = 0;

 81     mm2[0] = Begin ;

 82     HappyVal[mm[Begin]] = 0;

 83     for(i = 1 ; i < N ;i++)

 84     {

 85         scanf("%s%d",tem,&happy);

 86         if(strcmp("ROM",tem)==0) ROM = i;

 87         mm[tem] = i;

 88         mm2[i] = tem;

 89         HappyVal[i] = happy;

 90     }

 91 

 92     char x[4],y[4];

 93 

 94     for(i = 0 ; i < N ;i++)

 95     {

 96         for(j = 0 ; j < N ;j++)

 97         {

 98             Grap[i][j] = INF;

 99         }

100         d[i] = h[i] = INF;

101         pre[i] = -1;

102         Count[i] = 0;

103     }

104 

105     for(i = 0 ; i < K ;i++)

106     {

107         scanf("%s%s",x,y);

108         scanf("%d",&Grap[mm[x]][mm[y]]);

109         Grap[mm[y]][mm[x]] = Grap[mm[x]][mm[y]];

110     }

111 

112     Dijkstra( mm[Begin] , N);

113 

114     printf("%d %d %d %d\n",num[mm["ROM"]],d[mm["ROM"]],h[mm["ROM"]],h[mm["ROM"]]/Count[mm["ROM"]]);

115 

116     stack<int> ss;

117     i= mm["ROM"];

118     while(i != -1)

119     {

120         ss.push(i);

121         i = pre[i];

122     }

123     int fir = 1;

124     while(!ss.empty())

125     {

126         if(fir == 1)

127         {

128             fir = 0;

129             printf("%s",mm2[ss.top()].c_str());

130         }

131         else printf("->%s",mm2[ss.top()].c_str());

132         ss.pop();

133     }

134 

135     printf("\n");

136 

137     return 0;

138 }