hdu 3315 My Brute (二分图最佳匹配,KM算法)
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。
/* ***********************************************
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}