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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.
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
1 621 2 102 3 103 4 104 5 105 6 106 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 }
NUC_HomeWork1 -- POJ2067(最短路)
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