HDU 3315 My Brute(考虑优先次序的费用流）@

Seaco is a beautiful girl and likes play a game called “My Brute”. Before Valentine’s Day, starvae and xingxing ask seaco if she wants to spend the Valentine’s Day with them, but seaco only can spend it with one of them. It’s hard
to choose from the two excellent boys. So there will be a competition between starvae and xingxing. The competition is like the game “My Brute”. 



Now starvae have n brutes named from S1 to Sn and xingxing’s brutes are named from X1 to Xn. A competition consists of n games. At the beginning, starvae's brute Si must versus xingxing’s brute Xi. But it’s hard for starvae to win the competition, so starvae
can change his brutes’ order to win more games. For the starvae’s brute Si, if it wins the game, starvae can get Vi scores, but if it loses the game, starvae will lose Vi scores. Before the competition, starvae’s score is 0. Each brute can only play one game.
After n games, if starvae’s score is larger than 0, we say starvae win the competition, otherwise starvae lose it. 

It’s your time to help starvae change the brutes’ order to make starvae’s final score be the largest. If there are multiple orders, you should choose the one whose order changes the least from the original one. The original order is S1, S2, S3 … Sn-1, Sn, while
the final order is up to you. 

For starvae’s brute Si (maybe this brute is not the original brute Si, it is the ith brute after you ordered them) and xingxing’s brute Xi, at first Si has Hi HP and Xi has Pi HP, Si’s damage is Ai and Xi’s is Bi, in other words, if Si attacks, Xi will lose
Ai HP and if Xi attacks, Si will lose Bi HP, Si attacks first, then it’s Xi’s turn, then Si… until one of them’s HP is less than 0 or equal to 0, that, it lose the game, and the other win the game. 

Come on, starvae’s happiness is in your hand!

InputFirst line is a number n. (1<=n<=90) Then follows a line with n numbers mean V1 to Vn. (0<Vi<1000) Then follows a line with n numbers mean H1 to Hn. (1<=Hi<=100)Then follows a line with n numbers mean P1 to Pn. (1<=Pi<=100) Then follows a line
with n numbers mean A1 to An.(1<=Ai<=50) Then follows a line with n numbers mean B1 to Bn. (1<=Bi<=50) A zero signals the end of input and this test case is not to be processed.
OutputFor each test case, if starvae can win the competition, print the largest score starvae can get, and then follow a real percentage means the similarity between the original order and the final order you had changed, round it to three digits after
the decimal point. If starvae can’t win the competition after changing the order, please just print “Oh, I lose my dear seaco!” Maybe the sample can help you get it.
Sample Input

3
4 5 6
6 8 10
12 14 16
7 7 6
7 3 5
3
4 5 6
6 8 10
12 14 16
5 5 5
5 5 5
0

Sample Output

7 33.333%
Oh, I lose my dear seaco!

      这里用费用流再做一遍,首先我们求出任意Si与Xj决斗时,你能获得的分值Wij. 下面网络流建图:

       源点s编号0, S1到Sn编号1到n, X1到Xn编号n+1到2*n, 汇点t编号2*n+1.

       源点s到任意Si点有边 (s, i, 1, 0)

       任意Xi点到汇点t有边 (i+n, t, 1, 0)

       如果Si与Xj决斗的解过为Wij分值,那么有下面两种情况:

       i==j时, 有边(i ,j+n, 1, -Wij*(n+1)-1) (注意这里Wij取负数且乘以(n+1)且减一,取负数,是因为最终结果取反是你能获得的最大分数.减一是使得该原始决斗顺序能够得以保留.乘以(n+1)是因为把权值扩大n+1倍之后再+1最终的权值就算是+n了然后除以(n+1)还是能得到正真的分数值 )

       i!=j时,有边(i,j+n,1,-Wij*(n+1) )

       最终我们求最小费用的负数X即可. X%(n+1)就是我们保持原先决斗顺序的个数,X/(n+1)就是我们能获得的最终分数.，这题要注意最大值的给定，1<<30, 小了会超时

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 1e6+10;
const int M = 1000;
int head[M], cnt;
struct node
{
int from, to, cap, cost, next;
}p
;

void add(int u,int v,int w,int z)
{
p[cnt].from=u, p[cnt].to=v, p[cnt].cap=w, p[cnt].cost=z, p[cnt].next=head[u], head[u]=cnt++;
p[cnt].from=v, p[cnt].to=u, p[cnt].cap=0, p[cnt].cost=-z, p[cnt].next=head[v], head[v]=cnt++;
return ;
}
int d[M], pre[M], vis[M];
const int inf = 1<<30;
int min_cost_flow(int s,int t,int f)
{
int res=0;
while(f>0)
{
queue<int>q;
q.push(s);
for(int i=0;i<=t;i++) d[i]=inf;
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
vis[s]=1, d[s]=0;
while(!q.empty())
{
int u=q.front();q.pop();
for(int i=head[u];i!=-1;i=p[i].next)
{
int v=p[i].to;
if(d[v]>d[u]+p[i].cost&&p[i].cap>0)
{
d[v]=d[u]+p[i].cost;
pre[v]=i;
if(!vis[v])
{
q.push(v);
vis[v]=1;
}
}
}
vis[u]=0;
}
if(d[t]==inf) return 0;
int flow=f;
for(int i=pre[t];i!=-1;i=pre[p[i].from])
{
flow=min(flow,p[i].cap);
}
res+=flow*d[t];
for(int i=pre[t];i!=-1;i=pre[p[i].from])
{
p[i].cap-=flow, p[1^i].cap+=flow;
}
f-=flow;
}
return res;
}

int v
, h
, pi
, a
, b
;
int get(int x,int y)
{
int s1=h[x], s2=pi[y];
while(1)
{
s2-=a[x];
if(s2<=0) return v[x];
s1-=b[y];
if(s1<=0) return -v[x];
}
}

int main()
{
printf("%d\n",0x3f3f3f3f);
int n;
while(scanf("%d", &n)!=EOF&&n!=0)
{
memset(head,-1,sizeof(head));
cnt=0;
for(int i=1;i<=n;i++) scanf("%d", &v[i]);
for(int i=1;i<=n;i++) scanf("%d", &h[i]);
for(int i=1;i<=n;i++) scanf("%d", &pi[i]);
for(int i=1;i<=n;i++) scanf("%d", &a[i]);
for(int i=1;i<=n;i++) scanf("%d", &b[i]);
int s=0, t=2*n+2;
for(int i=1;i<=n;i++)
{
add(s,i,1,0);
add(n+i,t,1,0);
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
int s=get(i,j);
if(i==j) add(i,n+j,1,-s*(n+1)-1);
else add(i,n+j,1,-s*(n+1));
}
}
int x=-min_cost_flow(s,t,n);
int cnt1=(x%(n+1));
x/=(n+1);
double px=(1.0*cnt1)/(n);
if(x>0) printf("%d %.3lf%%\n",x,100*px);
else printf("Oh, I lose my dear seaco!\n");
}
return 0;
}