hdu 3315 My Brute (二分图最佳匹配,KM算法)

hdu3315 题目链接

题意:两个人分别各有n个怪物。进行n场pk.每只怪物必须恰好进行一场pk.如果先手的第i只怪物赢,会获得v[i]的val,输会减少v[i]的val.给出两个人n只怪物的血量和攻击力。先手的初始战斗顺序为1,2,3..n(后手的战斗顺序一直都是1,2,3..n) 现在问能否通过调整顺序使得先手获得的val最大,如果这个val大于0,表示先手可以赢。如果可以赢,那么还要求调整后的顺序和原始顺序的相似度,并且使得相似度尽可能大(If there are multiple orders, you should choose the one whose order changes the least from the original one)

思路:先根据血量和攻击力,n*n的时间处理出每场战斗的输赢信息,然后结合v[i],得到每两个怪物战斗的先手得到的val的值。

然后和hdu 1853类似,依然希望尽可能多的安排不改变。

我们的做法仍然是把w*N,然后钦定的w再+1

然后改变个数,由于存在负数。。。和hdu 1853的处理有区别。。。

想了一下。。。其实用link数组对照初始钦定顺序就好了啊。。。

1A.爽上天。。。

最近各种1A...?好开心。

选区_075

/* ***********************************************
Author :111qqz
Created Time :2016年06月03日 星期五 15时46分12秒
File Name :code/hdu/3315.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=105;
 7int n;
 8int val[N];
 9int h[N],p[N],a[N],b[N];
10int win[N][N];
11int w[N][N];
12int lx[N],ly[N];
13bool visx[N],visy[N];
14int link[N];
15int slk[N];
16int num;
 1bool find( int u)
 2{
 3    visx[u] = true;
 4    for ( int v = 1 ; v <= n ; v++)
 5    {
 6	if (visy[v]) continue;
 7	int tmp = lx[u] + ly[v] - w[u][v];
 8	if (tmp==0)
 9	{
10	    visy[v] = true;
1	    if (link[v]==-1||find(link[v]))
2	    {
3		link[v] = u;
4		return true;
5	    }
6	}else if (tmp<slk[v]) slk[v] = tmp;
7    }
    return false;
}
1int KM()
2{
3    ms(link,-1);
4    ms(lx,0xc0);
5    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    }
 1    for ( int i = 1 ; i <= n ; i++)
 2    {
 3	if (link[i]>-1)
 4	{
 5	    num = num + (link[i]==i);
 6	}
 7    }
 8   // cout<<"num:"<<num<<endl;
 9    //num%=N;
10  //  cout<<"num:"<<num<<endl;
11    int res = 0 ;
12    for ( int i = 1 ; i <= n ; i++)
13    {
14	if (link[i]>-1)
15	{
16	    if (w[link[i]][i]>0) res = res + w[link[i]][i]/N;
17	    else res = res + (w[link[i]][i]-2)/N; //???
18	}
19    }
    return res;
1}
2int main()
3{
4	#ifndef  ONLINE_JUDGE 
5	freopen("code/in.txt","r",stdin);
6  #endif
 1	while (scanf("%d",&n)!=EOF)
 2	{
 3	    if (n==0) break;
 4	    ms(w,0xc0);
 5	    ms(win,0);
 6	    num =  0;
 7	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&val[i]);
 8	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&h[i]);
 9	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&p[i]);
10	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
11	    for ( int i = 1 ; i <= n ; i++) scanf("%d",&b[i]);
1	    for ( int i = 1 ;i <= n ;i++)
2		for  (int j  = 1 ; j <= n ; j++)
3		{
4		    win[i][j] = (h[i]+b[j]-1)/b[j]+1>(p[j]+a[i]-1)/a[i]?1:-1;
5		}
1	    for ( int i = 1 ;i <= n ; i++)
2		for ( int j = 1 ;j  <= n ; j++)
3		    w[i][j] = win[i][j]*val[i]*N;
	    for ( int i = 1; i <= n ; i++) w[i][i]++;
 1	    int ans = KM();
 2	    if (ans<=0)
 3	    {
 4		puts("Oh, I lose my dear seaco!");
 5	    }
 6	    else
 7	    {
 8		double ret = num*1.0/(n*1.0);
 9		ret = ret * 100.0;
10		printf("%d %.3f%%\n",ans,ret);
11	    }
	}
1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}