HDU 3395 Special Fish（拆点+最大费用最大流）

Special Fish

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2367    Accepted Submission(s): 878

Problem Description

There is a kind of special fish in the East Lake where is closed to campus of Wuhan University. It’s hard to say which gender of those fish are, because every fish believes itself as a male, and it may attack one of some other fish who is believed to be female by it.
A fish will spawn after it has been attacked. Each fish can attack one other fish and can only be attacked once. No matter a fish is attacked or not, it can still try to attack another fish which is believed to be female by it.
There is a value we assigned to each fish and the spawns that two fish spawned also have a value which can be calculated by XOR operator through the value of its parents.
We want to know the maximum possibility of the sum of the spawns.

 

Input

The input consists of multiply test cases. The first line of each test case contains an integer n (0 < n <= 100), which is the number of the fish. The next line consists of n integers, indicating the value (0 < value <= 100) of each fish. The next n lines, each line contains n integers, represent a 01 matrix. The i-th fish believes the j-th fish is female if and only if the value in row i and column j if 1.
The last test case is followed by a zero, which means the end of the input.

 

Output

Output the value for each test in a single line.

 

Sample Input

3

1 2 3

011

101

110

 

0

 

Sample Output

6

 

题目链接：HDU 3395

这题跟卖啤酒那差不多，一条鱼只能攻击一次也只能被攻击一次，因此把鱼拆成攻击点1~n和被攻击点n+1~2*n，然后加入源汇点连边即可，但是最大的值不一定是最大流的情况下出现的，因此要在找dis[T]大于等于0的时候停止寻找即可。这题还有一个最大的坑！负号-和异或号^的优先级不一样，记得加括号…………

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=110;
const int M=N+N+N*N;
struct edge
{
    int to,nxt,cap,cost;
    edge(){}
    edge(int _to,int _nxt,int _cap,int _cost):to(_to),nxt(_nxt),cap(_cap),cost(_cost){}
};
edge E[M<<1];
int head[N<<1],tot;
int dis[N<<1],pre[N<<1],path[N<<1];
bitset<N<<1>vis;
char Mat[N][N];
int mc,mf,val[N];

void init()
{
    CLR(head,-1);
    tot=0;
    mc=mf=0;
}
inline void add(int s,int t,int cap,int cost)
{
    E[tot]=edge(t,head[s],cap,cost);
    head[s]=tot++;
    E[tot]=edge(s,head[t],0,-cost);
    head[t]=tot++;
}
int SPFA(int s,int t)
{
    CLR(dis,INF);
    vis.reset();
    queue<int>Q;
    dis[s]=0;
    vis[s]=1;
    Q.push(s);
    while (!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        vis[u]=0;
        for (int i=head[u]; ~i; i=E[i].nxt)
        {
            int v=E[i].to;
            if(dis[v]>dis[u]+E[i].cost&&E[i].cap>0)
            {
                dis[v]=dis[u]+E[i].cost;
                pre[v]=u;
                path[v]=i;
                if(!vis[v])
                {
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
    return dis[t]<0;
}
void MCMF(int s,int t)
{
    while (SPFA(s,t))
    {
        int Mf=INF;
        for (int i=t; i!=s; i=pre[i])
            Mf=min(Mf,E[path[i]].cap);
        for (int i=t; i!=s; i=pre[i])
        {
            E[path[i]].cap-=Mf;
            E[path[i]^1].cap+=Mf;
        }
        mf+=Mf;
        mc+=Mf*dis[t];
    }
}
int main(void)
{
    int n,i,j;
    while (~scanf("%d",&n)&&n)
    {
        init();
        for (i=1; i<=n; ++i)
            scanf("%d",&val[i]);
        int S=0,T=n+n+1;
        for (i=1; i<=n; ++i)
        {
            scanf("%s",Mat[i]+1);
            add(S,i,1,0);//n
            add(i+n,T,1,0);//n
            for (j=1; j<=n; ++j)
            {
                if(Mat[i][j]=='1')
                    add(i,n+j,1,-(val[i]^val[j]));//n*n
            }
        }
        MCMF(S,T);
        printf("%d
",-mc);
    }
    return 0;
}