【KM匹配】 HDOJ 2448 Mining Station on the Sea

最新推荐文章于 2018-07-21 12:05:43 发布

原创最新推荐文章于 2018-07-21 12:05:43 发布 · 425 阅读

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本文介绍了一种结合Floyd算法预处理最短路径与Kuhn-Munkres (KM) 算法解决匹配问题的方法。该方法适用于解决特定场景下的匹配问题，如港口只能被访问一次的情况。

先用floyd预处理出最短路，然后直接km匹配就可以了。。。注意预处理的时候由于只能到达港口一次，所以floyd要先枚举中间点，不能最后枚举中间点。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 305
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

int G[maxn][maxn];
int g[maxn][maxn];
int linker[maxn];
int slack[maxn];
bool visx[maxn];
bool visy[maxn];
int lx[maxn];
int ly[maxn];
int a[maxn];
int n, m, o, m1, m2, nx, ny;

bool dfs(int x)
{
	visx[x] = true;
	for(int y = 0; y < ny; y++) {
		if(visy[y]) continue;
		int tmp = lx[x] + ly[y] - G[x][y];
		if(!tmp) {
			visy[y] = true;
			if(linker[y] == -1 || dfs(linker[y])) {
				linker[y] = x;
				return true;
			}
		}
		else slack[y] = min(slack[y], tmp);
	}
	return false;
}

int km()
{
	memset(linker, -1, sizeof linker);
	memset(ly, 0, sizeof ly);
	for(int i = 0; i < nx; i++) {
		lx[i] = -INF;
		for(int j = 0; j < ny; j++)
			lx[i] = max(lx[i], G[i][j]);
	}
	for(int x = 0; x < nx; x++) {
		for(int i = 0; i < ny; i++)
			slack[i] = INF;
		while(true) {
			memset(visx, 0, sizeof visx);
			memset(visy, 0, sizeof visy);
			if(dfs(x)) break;
			int d = INF;
			for(int i = 0; i < ny; i++)
				if(!visy[i] && d > slack[i])
					d = slack[i];
			for(int i = 0; i < nx; i++)
				if(visx[i]) lx[i] -= d;
			for(int i = 0; i < ny; i++)
				if(visy[i]) ly[i] += d;
				else slack[i] -= d;
		}
	}
	int res = 0;
	for(int i = 0; i < ny; i++) res += G[linker[i]][i];
	return res;
}

void read()
{
	scanf("%d%d%d", &m, &m1, &m2);
	o = n + m;
	for(int i = 0; i < o; i++)
		for(int j = 0; j < o; j++)
			g[i][j] = INF;
	for(int i = 0; i < n; i++) scanf("%d", &a[i]), a[i]--, a[i] += n;
	int u, v, w;
	while(m1--) {
		scanf("%d%d%d", &u, &v, &w);
		u--, v--;
		u += n;
		v += n;
		g[u][v] = min(g[u][v], w);
		g[v][u] = min(g[v][u], w);
	}
	while(m2--) {
		scanf("%d%d%d", &u, &v, &w);
		u--, v--;
		v += n;
		g[u][v] = min(g[u][v], w);
		g[v][u] = min(g[v][u], w);
	}
}

void work()
{
	for(int k = n; k < o; k++)
		for(int i = 0; i < o; i++)
			for(int j = 0; j < o; j++)
				g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++)
			if(g[a[i]][j] == INF) G[i][j] = 0;
			else G[i][j] = -g[a[i]][j];
	nx = ny = n;
	printf("%d\n", -km());
}

int main()
{
	while(scanf("%d", &n)!=EOF) {
		read();
		work();
	}
	
	
	return 0;
}