本文主要是介绍HDU 3435 A new Graph Game（KM完美匹配），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
HDU 3435 A new Graph Game
题目链接
题意：又是这类求环总和的最小值，一个点只能在一个环上
思路：KM完美匹配求解，不过这题点有1000，理论上n^3算法是过不了的，不过也没有更好的方法了，用费用流来做的话也是n^3，而且常数还更大
代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;const int MAXNODE = 1005;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();}}int ans = 0;for (int i = 0; i < n; i++) {if (g[left[i]][i] == -INF) {printf("NO\n");return;}ans += g[left[i]][i];}printf("%d\n", -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, w;while (m--) {scanf("%d%d%d", &u, &v, &w);u--; v--;gao.add_Edge(u, v, -w);gao.add_Edge(v, u, -w);}printf("Case %d: ", ++cas);gao.km();}return 0;
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