1072 Gas Station (30point(s))
A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range.
Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 positive integers: N (≤103), the total number of houses; M (≤10), the total number of the candidate locations for the gas stations; K (≤104), the number of roads connecting the houses and the gas stations; and DS, the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.
Then K lines follow, each describes a road in the format
P1 P2 Dist
where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.
Output Specification:
For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output No Solution.
Sample Input:
4 3 11 5
1 2 2
1 4 2
1 G1 4
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2
Sample Output:
G1
2.0 3.3
Sample Input:
2 1 2 10
1 G1 9
2 G1 20
Sample Output:
No Solution
題目大意：
輸入 N 個村莊，M 個加油站，K 條路徑，DS 一個加油站服務的範圍
從 M 個加油站中求一個站點，符合離居民區最遠，並且服務區 DS 依舊覆蓋居民區，若有多個，則輸出距離所有居民區平均距離最小的，若還有多個，則輸出編號最小的
設計思路：
Dijkstra 計算最短路徑
- 對 M 個加油站依次計算 Dijkstra 最短路徑
- 判斷此加油站的服務範圍覆蓋，最近村莊最遠，平均距離最小，編號最小
- 輸出最終編號
編譯器：C (gcc)
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int c2i(char *a, int n)
{
if (a[0] == 'G')
return atoi(a + 1) + n;
return
atoi(a);
}
int e[1020][1020] = {0};
int dis[1020], visit[1020];
void dijkstra(int start, int n)
{
int i, j, u, v, min;
for (j = 0; j <= n; j++) {
dis[j] = INT_MAX;
visit[j] = 0;
}
dis[start] = 0;
for (i = 0; i < n; i++) {
u = -1;
min = INT_MAX;
for (j = 1; j <= n; j++) {
if (visit[j] == 0 && dis[j] < min) {
u = j;
min = dis[j];
}
}
if (u == -1)
break;
visit[u] = 1;
for (v = 1; v <= n; v++)
if (!visit[v] && e[u][v] && dis[u] + e[u][v] < dis[v])
dis[v] = dis[u] + e[u][v];
}
}
int main(void)
{
int n, m, k, ds, station = -1;
double mindis = -1.0, avgdis = INT_MAX * 1.0, mindis_now, avgdis_now;
char s1[5], s2[5];
int i, j, u, v, flag;
scanf("%d%d%d%d", &n, &m, &k, &ds);
for (i = 0; i < k; i++) {
scanf("%s%s%d", s1, s2, &j);
u = c2i(s1, n);
v = c2i(s2, n);
e[u][v] = j;
e[v][u] = j;
}
for (i = n + 1; i <= n + m; i++) {
dijkstra(i, n + m);
mindis_now = INT_MAX * 1.0;
avgdis_now = 0.0;
flag = 1;
for (j = 1; j <= n && flag; j++) {
if (dis[j] <= ds) {
mindis_now = mindis_now < dis[j] * 1.0 ? mindis_now : dis[j] * 1.0;
avgdis_now += dis[j] * 1.0;
} else {
flag = 0;
}
}
avgdis_now /= n;
if (flag && (mindis_now > mindis || (mindis_now == mindis && avgdis_now < avgdis))) {
station = i;
mindis = mindis_now;
avgdis = avgdis_now;
}
}
if (station != -1)
printf("G%d\n%.1lf %.1lf", station - n, mindis, avgdis);
else
printf("No Solution");
return 0;
}