题目链接:http://cogs.pro:8081/cogs/problem/problem.php?pid=FSXJmiJSg
问题描述
为了进一步普及九年义务教育,政府要在某乡镇建立P所希望小学,该乡镇共有n个村庄,村庄间的距离已知,请问学校建在哪P个村庄最好?(好坏的标准是学生就近入学,即在来上学的学生中,以最远的学生走的路程为标准。或者说最远的学生与学校的距离尽可能的小。)
#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #define MAXN 110 using namespace std; int edge[MAXN][MAXN]; int res[MAXN]; int temp[MAXN]; //临时数组 int n, m, p, maxDis, minDis = 99999999, ans = 99999999; void dfs(int x, int now) //x表示往数组里加了几个,now表示加的点 { if (x == p + 1) { maxDis = 0; for (int i = 0; i < n; i++) { minDis = 99999999; for (int j = 1; j <= p; j++) { minDis = min(minDis, edge[i][temp[j]]); //找到村庄i与p个学校最近的那个学校的距离 } maxDis = max(maxDis, minDis); //找到以后更新最远的距离 } if (ans > maxDis) //最远的学生走的路最少 { ans = maxDis; memset(res, 0, sizeof(res)); for (int i = 0; i < n; i++) res[i] = temp[i]; } return; } for (int i = now + 1; i < n; i++) { temp[x] = i; dfs(x + 1, i); } } int main() { freopen("djsc.in", "r", stdin); freopen("djsc.out", "w", stdout); int x, y, v; scanf("%d %d %d", &n, &m, &p); for(int i = 0; i < n; i++) for (int j = 0; j < n; j++) { edge[i][j] = 99999999; edge[i][i] = 0; } for (int i = 0; i < m; i++) { scanf("%d%d%d", &x, &y, &v); edge[x][y] = edge[y][x] = v; } for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) edge[i][j] = min(edge[i][j], edge[i][k] + edge[k][j]); dfs(1, -1); for (int i = 1; i <= p; i++) printf("%d ", res[i]); return 0; }
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