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;
}