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

Posted by 111qqz on Thursday, June 2, 2016

TOC

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
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=205;
int n,m,k,p;
int id[N];
int a[N][N],b[N][N];
int lx[N],ly[N];
int link[N];
bool visx[N],visy[N];
int slk[N];
int w[N][N];

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

    for ( int v = 1 ; v <= n ; v++)
    {
    if (visy[v]) continue;
    int tmp = lx[u] + ly[v] - w[u][v];
    if (tmp==0)
    {
        visy[v] = true;
        if (link[v]==-1||find(link[v]))
        {
        link[v] = u;
        return true;
        }
    }else if(tmp<slk[v]) slk[v] = tmp;
    }
    return false;
}
LL KM()
{
    ms(link,-1);
    ms(lx,0xc0);
    ms(ly,0);

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


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

    while (1)
    {
        ms(visx,false);
        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];

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

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

    return -res;
    
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif

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

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

        for ( int kk = 1 ; kk <= m ; kk++)
        for ( int i = 1 ; i <= m ; i++)
            for ( int j = 1 ; j <= m ; j++)
            a[i][j] = min(a[i][j],a[i][kk]+a[kk][j]);
        for ( int i = 1 ; i <= p ; i++)
        {
        int u,v,cost;
        scanf("%d%d%d",&u,&v,&cost);
        for ( int j = 1 ; j <= n ; j++)
            w[j][u] = min(w[j][u],a[id[j]][v]+cost); 
        }


//	    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。
//		for ( int i = 1 ; i <= n ; i++)
//		    for ( int j = 1 ; j <= m ; j++)
//			if (b[i][kk]+a[kk][j]<b[i][j]) b[i][j] = b[i][kk] + a[kk][j];

        for ( int i = 1 ; i <= n ; i++)
        for ( int j = 1 ; j <= n ; j++)
             w[i][j]=-w[i][j];
	    
        LL ans = KM();
        printf("%lld\n",ans);


    }

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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