hdu 2448 Mining Station on the Sea (floyd+KM)

2016-06-02 · 2 min read · floyd KM 二分图最佳匹配

hdu2448 题目链接

题意：n个船n个港口，一个港口只能承接一个船，m个油田，给出n个船各自在哪个油田，然后给出m个油田之间的无相图，然后给出油田和港口之间的有向图。求n个船到达港口的最小距离之和。

思路：想到了用floyd先更新一下距离，然后KM.不过思维不够严谨，只更新了港口通过油田到达油田的距离，而没有更新油田通过油田到达油田的距离QAQ.

所以应该先更新油田通过油田到达油田的距离，然后再更新港口通过油田到达油田的距离。。。

哦，还有。。。不要把n个船所对应的港口作为下标。。而是转化成1..n，这样写KM里会比较好写。。。不然总得带着那个id[i].

选区_074

/* ***********************************************
Author :111qqz
Created Time :2016年06月03日 星期五 03时07分16秒
File Name :code/hdu/2448.cpp
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=205;
 7int n,m,k,p;
 8int id[N];
 9int a[N][N],b[N][N];
10int lx[N],ly[N];
11int link[N];
12bool visx[N],visy[N];
13int slk[N];
14int w[N][N];

1bool find( int u)
2{
3 //   cout<<"u:"<<u<<endl;
4    visx[u] = true;

 1    for ( int v = 1 ; v <= n ; v++)
 2    {
 3	if (visy[v]) continue;
 4	int tmp = lx[u] + ly[v] - w[u][v];
 5	if (tmp==0)
 6	{
 7	    visy[v] = true;
 8	    if (link[v]==-1||find(link[v]))
 9	    {
10		link[v] = u;
11		return true;
12	    }
13	}else if(tmp<slk[v]) slk[v] = tmp;
14    }
15    return false;
16}
17LL KM()
18{
19    ms(link,-1);
20    ms(lx,0xc0);
21    ms(ly,0);

1    for ( int i = 1 ; i <= n ; i++)
2	for ( int j =  1 ; j <= n ; j++)
3	    lx[i] = max(lx[i],w[i][j]);

1    for ( int i = 1 ; i <= n ; i++)
2    {
3	ms(slk,0x3f);

1	while (1)
2	{
3	    ms(visx,false);
4	    ms(visy,false);

	    if (find(i)) break;

	    int d = inf;

	    for ( int j = 1 ; j <= n ; j++)
		if (!visy[j]&&slk[j]<d) d = slk[j];

1	    for ( int j = 1 ; j <= n ; j++)
2		if (visx[j]) lx[j]-=d;
3	    for ( int j = 1 ; j <= n ; j++) 
4		if (visy[j]) ly[j]+=d ; else slk[j]-=d;
5	}
6    }
7    LL res = 0LL ;

    for ( int i= 1 ; i <= n ; i++)
	if (link[i]>-1) res += LL(w[link[i]][i]);

    return -res;

1}
2int main()
3{
4	#ifndef  ONLINE_JUDGE 
5	freopen("code/in.txt","r",stdin);
6  #endif

1	while (scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF)
2	{
3	    ms(a,0x3f);
4	    ms(b,0x3f);
5	    ms(id,0);
6	    ms(w,0x3f);
7	    for ( int i = 1 ;i <= m ; i++) a[i][i] = 0 ;
8	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&id[i]);

1	    for ( int i = 1 ; i <= k ; i++)
2	    {
3		int u,v,cost;
4		scanf("%d%d%d",&u,&v,&cost);
5		//a[u][v] = max(a[u][v],-w);
6		a[u][v] = min(a[u][v],cost);
7		a[v][u] = min(a[v][u],cost);
8	    }

 1	    for ( int kk = 1 ; kk <= m ; kk++)
 2		for ( int i = 1 ; i <= m ; i++)
 3		    for ( int j = 1 ; j <= m ; j++)
 4			a[i][j] = min(a[i][j],a[i][kk]+a[kk][j]);
 5	    for ( int i = 1 ; i <= p ; i++)
 6	    {
 7		int u,v,cost;
 8		scanf("%d%d%d",&u,&v,&cost);
 9		for ( int j = 1 ; j <= n ; j++)
10		    w[j][u] = min(w[j][u],a[id[j]][v]+cost); 
11	    }

1//	    for ( int kk = 1 ; kk <= m ; kk++)  //这只是港口经过中间station到达station的距离近了，还有station经过中间station到达station距离近的情况没考虑,必须放在一起更新。
2//		for ( int i = 1 ; i <= n ; i++)
3//		    for ( int j = 1 ; j <= m ; j++)
4//			if (b[i][kk]+a[kk][j]<b[i][j]) b[i][j] = b[i][kk] + a[kk][j];

1	    for ( int i = 1 ; i <= n ; i++)
2		for ( int j = 1 ; j <= n ; j++)
3		     w[i][j]=-w[i][j];

	    LL ans = KM();
	    printf("%lld\n",ans);


	}

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}