NUC_HomeWork1 -- POJ2067(最短路)

C - Fire Station

Description

A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose the location of the fire station so as to reduce the distance to the nearest station from the houses of the disgruntled residents. 
The city has up to 500 intersections, connected by road segments of various lengths. No more than 20 road segments intersect at a given intersection. The location of houses and firestations alike are considered to be at intersections (the travel distance from the intersection to the actual building can be discounted). Furthermore, we assume that there is at least one house associated with every intersection. There may be more than one firestation per intersection. 

Input

The first line of input contains two positive integers: f,the number of existing fire stations (f <= 100) and i, the number of intersections (i <= 500). The intersections are numbered from 1 to i consecutively. f lines follow; each contains the intersection number at which an existing fire station is found. A number of lines follow, each containing three positive integers: the number of an intersection, the number of a different intersection, and the length of the road segment connecting the intersections. All road segments are two-way (at least as far as fire engines are concerned), and there will exist a route between any pair of intersections.

Output

You are to output a single integer: the lowest intersection number at which a new fire station should be built so as to minimize the maximum distance from any intersection to the nearest fire station.

Sample Input

1 6

2

1 2 10

2 3 10

3 4 10

4 5 10

5 6 10

6 1 10

Sample Output

5
这是仿照人家写的SPFA，然后补上其他3个最短路算法

  1 /好长时间没写过最短路了，好多基础的东西都不记得了  c

  2 #include <cstdio>

  3 #include <iostream>

  4 #include <cstring>

  5 #include <cmath>

  6 #include <algorithm>

  7 #include <queue>

  8 

  9 using namespace std;

 10 

 11 

 12 const int N = 5050, INF = 10000;

 13 

 14 int head[N], nc, n; ///

 15 int dis[N];         ///计算距离的数组

 16 int stk[N], f, r;   ///用数组模拟栈，

 17 bool vis[N];        ///在用spfa时的标记数组

 18 

 19 struct Edge{        ///边

 20     int to, next, cost;

 21 }edge[100000];

 22 

 23 void add(int a, int b, int c)  ///加边函数，双向

 24 {

 25     edge[nc].to = b;

 26     edge[nc].next = head[a];

 27     edge[nc].cost = c;

 28     head[a] = nc++;

 29 

 30     edge[nc].to = a;

 31     edge[nc].next = head[b];

 32     edge[nc].cost = c;

 33     head[b] = nc++;

 34 }

 35 

 36 void fire_spfa(int fire[], int num)

 37 {

 38     memset(vis, false, sizeof(vis));

 39 

 40     f = r = 0;                      ///初始化栈

 41 

 42     for(int i = 0; i < num; i++)    ///初始化每个消防站到其他点的距离

 43     {

 44         int t = fire[i];

 45         if(!vis[t])

 46         {

 47             vis[t] = true;

 48             dis[stk[r++] = t] = 0;///初始化，并将当前点入栈

 49         }

 50     }

 51 

 52     while(f != r)               ///如果栈不为空

 53     {

 54         int now = stk[f++];     ///从栈中取出栈顶元素

 55         if(f == N) f = 0;       ///防止栈空间不够

 56 

 57         vis[now] = false;       ///将当前结点标记

 58         for(int i = head[now]; i != -1; i = edge[i].next)   ///从边表中取出元素

 59         {

 60             int to = edge[i].to;

 61             int cot = edge[i].cost;

 62 

 63             if(dis[to] > dis[now] + cot)                    ///进行松弛操作

 64             {

 65                 dis[to] = dis[now] + cot;

 66                 if(!vis[to])                                ///如果前驱没有访问过，

 67                 {

 68                     stk[r++] = to;                          ///入栈

 69                     if(r == N)

 70                         r = 0;

 71                     vis[to] = true;                         ///标记

 72                 }

 73             }

 74         }

 75     }

 76 }

 77 

 78 int spfa(int src)

 79 {

 80     memset(vis, false, sizeof(vis));

 81     f = 0;

 82     r = 1;

 83 

 84     vis[src] = true;

 85     int d[N];

 86     for(int i = 1; i <= n; ++i)     ///从当前结点到其余所有结点距离初始化

 87     {

 88         d[i] = INF;

 89     }

 90 

 91     d[src] = 0;

 92     stk[0] = src;

 93 

 94     while(f != r)

 95     {

 96         int now = stk[f++];

 97         if(f == N)

 98             f = 0;

 99         vis[now] = false;

100 

101         for(int i = head[now]; i != -1; i = edge[i].next)

102         {

103             int to = edge[i]. to;

104             int cot = edge[i].cost;

105 

106             if(d[to] > d[now] + cot)

107             {

108                 d[to] = d[now] + cot;

109                 if(!vis[to])

110                 {

111                     stk[r++] = to;

112                     vis[to] = true;

113                     if(r == N)

114                         r = 0;

115                 }

116             }

117         }

118     }

119 

120     int ans = 0;

121     for(int i = 1; i <= n; ++i)

122     {

123         ans = max(ans, min(d[i], dis[i]));

124     }

125     return ans;

126 }

127 

128 int main()

129 {

130     int ff, fire[N], a, b, c;

131 

132     memset(head, -1, sizeof(head));

133     nc = 0;

134     scanf("%d%d", &ff, &n);

135     for(int i = 0; i < ff; ++i)

136     {

137         scanf("%d", fire+i);

138     }

139 

140     while(scanf("%d%d%d", &a, &b, &c) != EOF)

141     {

142         add(a, b, c);

143     }

144 

145     for(int i = 1; i <= n; ++i)

146         dis[i] = INF;

147 

148     fire_spfa(fire, ff);

149 

150     int id = 1, ans = INF;

151 

152     for(int i = 1; i <= n; ++i)

153     {

154         if(dis[i] != 0)

155         {

156             int tp = spfa(i);

157             if(ans > tp)

158             {

159                 ans = tp;

160                 id = i;

161             }

162         }

163     }

164 

165     printf("%d\n", id);

166     return 0;

167 }

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