做一个最短路 一个最长路，枚举两点求比值最大的记录，然后最后输出就行
#include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long LL; const double PI = acos(-1.0); template inline T MAX(T a, T b){if (a > b) return a;return b;} template inline T MIN(T a, T b){if (a < b) return a;return b;} const int N = 111; const int M = 11111; const LL MOD = 1000000007LL; const int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1}; const int INF = 0x3f3f3f3f; int dist[1111][1111]; int down[1111][1111]; int path1[1111][1111]; int path2[1111][1111]; struct node { int v, w, next; }edge[2222], redge[2222]; int head1[1111], head2[1111], cnt1, cnt2; int n, m1, m2, inq[1111]; inline void addnode1(int u, int v, int w) { edge[cnt1].v = v; edge[cnt1].w = w; edge[cnt1].next = head1[u]; head1[u] = cnt1++; } inline void addnode2(int u, int v, int w) { redge[cnt2].v = v; redge[cnt2].w = w; redge[cnt2].next = head2[u]; head2[u] = cnt2++; } void spfa1(int d[], int v0) { fill(d, d + n + 1, -1); memset(inq, 0, sizeof(inq)); queue < int > q; q.push(v0); inq[v0] = 1; d[v0] = 0; path1[v0][v0] = -1; int u, v, i, j, k, w; while (!q.empty()) { u = q.front(); q.pop(); inq[u]--; for (i = head1[u]; ~i; i = edge[i].next) { v = edge[i].v; w = edge[i].w; // if (v == v0) continue; if (d[u] + w > d[v]) { d[v] = d[u] + w; path1[v0][v] = u; if (!inq[v]) {inq[v]++; q.push(v);} } } } } void spfa2(int d[], int v0) { fill(d, d + n + 1, INF); memset(inq, 0, sizeof(inq)); queue < int > q; q.push(v0); path2[v0][v0] = -1; inq[v0] = 1; d[v0] = 0; int u, v, i, j, k, w; while (!q.empty()) { u = q.front(); q.pop(); inq[u]--; for (i = head2[u]; ~i; i = redge[i].next) { v = redge[i].v; w = redge[i].w; // if (v == v0) continue; if (d[u] + w < d[v]) { d[v] = d[u] + w; path2[v0][v] = u; if (!inq[v]) {inq[v]++; q.push(v);} } } } } void output1(int p, int k) { if (path2[p][k] != -1) { output1(p, path2[p][k]); } printf("%d ", k); } void output2(int p, int k) { if (path1[p][path1[p][k]] != -1) { output2(p, path1[p][k]); } if (path1[p][path1[p][k]] == -1) printf("%d", k); else printf(" %d", k); } int main() { int T; scanf("%d", &T); while (T--) { int i,j ,k, u, v, w; memset(head1, -1, sizeof(head1)); cnt1 = cnt2 = 0; memset(head2, -1, sizeof(head2)); scanf("%d%d%d", &n, &m1, &m2); for (i = 0; i < m1; ++i) { scanf("%d%d%d", &u, &v, &w); addnode1(u, v, w); } for (i = 0; i < m2; ++i) { scanf("%d%d%d", &u, &v, &w); addnode2(u, v, w); } for (i = 1; i <= n; ++i) { spfa1(down[i], i); spfa2(dist[i], i); } int pos1, pos2; double mm = 0; for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j) { if (down[j][i] != -1 && dist[i][j] != INF && i != j) { if (1.0 * (down[j][i]) / dist[i][j] > mm) { pos1 = i; pos2 = j; mm = 1.0 * (down[j][i]) / dist[i][j]; } } } // for (i = 1; i <= n; ++i) // { // for (j = 1; j <= n; ++j) // printf("%d ", dist[i][j]); // printf("\n"); // } output1(pos1, pos2); output2(pos2, pos1); printf("\n%.3f\n", mm); } return 0; }
