标签:des style http color io os ar for sp
题意:给定一些有向带权边,求出把这些边构造成一个个环,总权值最小
思路:由于环入度出度为1,所以可以把每个点拆成入度点和出度点,然后建图做一次二分图完美匹配即可,注意这题坑点,有重边
代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MAXNODE = 105; typedef int Type; const Type INF = 0x3f3f3f3f; struct KM { int n; Type g[MAXNODE][MAXNODE]; Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE]; int left[MAXNODE]; bool S[MAXNODE], T[MAXNODE]; void init(int n) { this->n = n; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { g[i][j] = -INF; } } } void add_Edge(int u, int v, Type val) { g[u][v] = max(g[u][v], val); } bool dfs(int i) { S[i] = true; for (int j = 0; j < n; j++) { if (T[j]) continue; Type tmp = Lx[i] + Ly[j] - g[i][j]; if (!tmp) { T[j] = true; if (left[j] == -1 || dfs(left[j])) { left[j] = i; return true; } } else slack[j] = min(slack[j], tmp); } return false; } void update() { Type a = INF; for (int i = 0; i < n; i++) if (!T[i]) a = min(a, slack[i]); for (int i = 0; i < n; i++) { if (S[i]) Lx[i] -= a; if (T[i]) Ly[i] += a; } } void km() { for (int i = 0; i < n; i++) { left[i] = -1; Lx[i] = -INF; Ly[i] = 0; for (int j = 0; j < n; j++) Lx[i] = max(Lx[i], g[i][j]); } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) slack[j] = INF; while (1) { for (int j = 0; j < n; j++) S[j] = T[j] = false; if (dfs(i)) break; else update(); } } } void solve() { km(); int ans = 0, flag = 0; for (int i = 0; i < n; i++) { if (g[left[i]][i] == -INF) { flag = 1; break; } ans += g[left[i]][i]; } if (flag) printf("N\n"); else printf("%d\n", -ans); } } gao; int n; int main() { while (~scanf("%d", &n) && n) { gao.init(n); int a, b; for (int i = 0; i < n; i ++) { while (scanf("%d", &a)) { if (a == 0) break; scanf("%d", &b); a--; gao.add_Edge(i, a, -b); } } gao.solve(); } return 0; }
UVA 1349 - Optimal Bus Route Design(KM完美匹配)
标签:des style http color io os ar for sp
原文地址:http://blog.csdn.net/accelerator_/article/details/39547309