UVA 11183 - Teen Girl Squad
题意:本质就是给一个有向图,有一些能连接的边,连接有一个代价,问从0能遍历到所有点的,并且代价最小的最小代价
思路:最小树形图的裸题,用朱刘算法求解,验证一下模板
代码:
#include <cstdio> #include <cstring> const int MAXNODE = 1005; const int MAXEDGE = 40005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type dist; Edge() {} Edge(int u, int v, Type dist) { this->u = u; this->v = v; this->dist = dist; } }; struct Directed_MST { int n, m; Edge edges[MAXEDGE]; int vis[MAXNODE]; int pre[MAXNODE]; int id[MAXNODE]; Type in[MAXNODE]; void init(int n) { this->n = n; m = 0; } void add_Edge(int u, int v, Type dist) { edges[m++] = Edge(u, v, dist); } void add_Edge(Edge e) { edges[m++] = e; } Type dir_mst(int root) { Type ans = 0; while (true) { for (int i = 0; i < n; i++) in[i] = INF; for (int i = 0; i < m; i++) { int u = edges[i].u; int v = edges[i].v; if (edges[i].dist < in[v] && u != v) { in[v] = edges[i].dist; pre[v] = u; } } for (int i = 0; i < n; i++) { if (i == root) continue; if (in[i] == INF) return -1; } int cnt = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for (int i = 0; i < n; i++) { ans += in[i]; int v = i; while (vis[v] != i && id[v] == -1 && v != root) { vis[v] = i; v = pre[v]; } if (v != root && id[v] == -1) { for (int u = pre[v]; u != v; u = pre[u]) id[u] = cnt; id[v] = cnt++; } } if (cnt == 0) break; for (int i = 0; i < n; i++) if (id[i] == -1) id[i] = cnt++; for (int i = 0; i < m; i++) { int v = edges[i].v; edges[i].u = id[edges[i].u]; edges[i].v = id[edges[i].v]; if (edges[i].u != edges[i].v) edges[i].dist -= in[v]; } n = cnt; root = id[root]; } return ans; } } gao; int t, n, m; int main() { int cas = 0; scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); gao.init(n); int u, v, d; while (m--) { scanf("%d%d%d", &u, &v, &d); gao.add_Edge(u, v, d); } int tmp = gao.dir_mst(0); printf("Case #%d: ", ++cas); if (tmp == -1) printf("Possums!\n"); else printf("%d\n", tmp); } return 0; }
时间: 2024-10-02 15:51:57