USACO2.4.3 Cow Tours (cowtour)

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注意题目中的直径并不是指距离最远的两点的距离，而是最短路最长两点的最短路。
用map[i][j]表示两点之间的最短路。
读入时初始化map，如果两点之间有边则map[i][j]=dist(i, j)否则map[i][j]=INF.
然后用一次floyd算出任意两点最短路。
mmax[i]表示从节点i出发，能到达的最远的节点的路程。
对于每一个不联通的i，j，将他俩接通，此时直径为mmax[i]+mmax[j]+dist(i,j).

枚举每一个不联通的i，j，计算直径统计最大值。

/*ID:shijiey1PROG:cowtourLANG:C++*/#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define INF 0x3f3f3f3fusing namespace std;int n;int x[155], y[155];double map[155][155];double mmax[155];double res = INF;double dist(int a, int b) {return sqrt((x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]));}int main() {freopen("cowtour.in", "r", stdin);freopen("cowtour.out", "w", stdout);memset(mmax, 0, sizeof(mmax));scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d %d\n", &x[i], &y[i]);char c;for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {scanf("%c", &c);if (c == '1') map[i][j] = dist(i, j);else map[i][j] = INF;if (i == j) map[i][j] = 0;}getchar();}for (int k = 1; k <= n; k++)for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)map[i][j] = min(map[i][j], map[i][k] + map[k][j]);for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (map[i][j] != INF)mmax[i] = max(mmax[i], map[i][j]);for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (map[i][j] == INF)res = min(res, mmax[i] + mmax[j] + dist(i, j));for (int i = 1; i <= n; i++)res = max(res, mmax[i]);printf("%lf\n", res);return 0;}

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