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HDU 2448 Mining Station on the Sea(KM最大匹配+floyd)

2018年10月11日 ⁄ 综合 ⁄ 共 1943字 ⁄ 字号 评论关闭

HDU 2448 Mining Station on the Sea

题目链接

题意:给定n个港口,m个油田,k行表示油田之间连边关系,p行表示港口和油田之间连边关系,都是无向边,在给定n个船起始在的油田位置,求每个船要开进一个港口,最短的总距离。注意,港口只能进入一次,进入就不能出来了

思路:先把油田和港口都连边,利用floyd求出最短路,注意由于港口不能做中转,所以枚举的时候不以港口作为中间点,然后每个船和每个港口距离连边,求KM最大匹配即可

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, m;
	Type g[MAXNODE][MAXNODE];
	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
	int left[MAXNODE], right[MAXNODE];
	bool S[MAXNODE], T[MAXNODE];

	void init(int n, int m) {
		this->n = n;
		this->m = m;
	}

	void add_Edge(int u, int v, Type val) {
		g[u][v] = val;
	}

	bool dfs(int i) {
		S[i] = true;
		for (int j = 0; j < m; 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;
					right[i] = j;
					return true;
				}
			} else slack[j] = min(slack[j], tmp);
		}
		return false;
	}

	void update() {
		Type a = INF;
		for (int i = 0; i < m; i++)
			if (!T[i]) a = min(a, slack[i]);
		for (int i = 0; i < n; i++)
			if (S[i]) Lx[i] -= a;
		for (int i = 0; i < m; i++)
			if (T[i]) Ly[i] += a;
	}

	Type km() {
		memset(left, -1, sizeof(left));
		memset(right, -1, sizeof(right));
		memset(Ly, 0, sizeof(Ly));
		for (int i = 0; i < n; i++) {
			Lx[i] = -INF;
			for (int j = 0; j < m; j++)
				Lx[i] = max(Lx[i], g[i][j]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) slack[j] = INF;
			while (1) {
				memset(S, false, sizeof(S));
				memset(T, false, sizeof(T));
				if (dfs(i)) break;
				else update();
			}
		}
		Type ans = 0;
		for (int i = 0; i < n; i++)
			ans += g[i][right[i]];
		return -ans;
	}
} gao;

const int N = 305;

int n, m, k, p;
int st[N], g[N][N];

int main() {
	while (~scanf("%d%d%d%d", &n, &m, &k, &p)) {
		for (int i = 0; i < n; i++) {
			scanf("%d", &st[i]);
			st[i]--;
		}
		int u, v, w;
		int tot = n + m;
		for (int i = 0; i < tot; i++)
			for (int j = 0; j < tot; j++) {
				if (i == j) g[i][j] = 0;
				else g[i][j] = INF;
			}
		while (k--) {
			scanf("%d%d%d", &u, &v, &w);
			u--; v--;
			g[u][v] = g[v][u] = min(g[u][v], w);
		}
		while (p--) {
			scanf("%d%d%d", &u, &v, &w);
			u--; v--;
			g[u + m][v] = g[v][u + m] = min(g[u + m][v], w);
		}
		for (int k = 0; k < m; k++)
			for (int i = 0; i < tot; i++)
				for (int j = 0; j < tot; j++)
					g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
		gao.init(n, n);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				gao.add_Edge(i, j, -g[st[i]][j + m]);
			}
		}
		printf("%d\n", gao.km());
	}
	return 0;
}

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