HDU 3395 Special Fish

以前通過KM來解的，今天用最大費用最大流寫一次，結果發現WA。。次奧，這不科學。。。後來才發現有一種情況沒有考慮到，即首先滿足的應該是最大費用，而不是流。。。

這兒有兩種解決方法：

1、一種是在加邊，即把二分圖左邊的魚與匯點相連，流量爲1，費用爲0；這樣才能保證在最大費用時，流量也是最大的。傳送門

2、直接做最小費用流（不是最小費用最大流），將增廣的結束條件改爲d[t]>=0即可。如果網絡負費用圈，則需要消圈。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 1010;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int from, to, cap, flow, cost;
	Edge(int from, int to, int cap, int flow, int cost): from(from), to(to), cap(cap), flow(flow), cost(cost) {}
};

struct MCMF
{
	int n, m, s, t;
	vector<Edge> edges;
	vector<int> G[maxn];
	int inq[maxn];
	int d[maxn];
	int p[maxn];
	int a[maxn];
	
	void init(int n)
	{
		this->n = n;
		for(int i = 0; i <= n; i++) G[i].clear();
		edges.clear();
	}
	
	void AddEdge(int from, int to, int cap, int cost)
	{
		edges.push_back(Edge (from, to, cap, 0, cost));
		edges.push_back(Edge (to, from, 0, 0, -cost));
		m = edges.size();
		G[from].push_back(m-2);
		G[to].push_back(m-1);
	}
	
	bool spfa(int s, int t, int &flow, int &cost)
	{
		for(int i = 0; i <= n; i++) d[i] = INF;
		memset(inq, 0, sizeof(inq));
		d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
		
		queue<int> Q;
		Q.push(s);
		while(!Q.empty())
		{
			int u = Q.front(); Q.pop();
			inq[u] = 0;
			for(int i = 0; i < G[u].size(); i++)
			{
				Edge &e = edges[G[u][i]];
				if(e.cap > e.flow && d[e.to] > d[u]+e.cost)
				{
					d[e.to] = d[u]+e.cost;
					p[e.to] = G[u][i];
					a[e.to] = min(a[u], e.cap-e.flow);
					if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
				}
			}
		}
		if(d[t] == INF) return 0;
		flow += a[t];
		cost += d[t]*a[t];
		int u = t;
		while(u != s)
		{
			edges[p[u]].flow += a[t];
			edges[p[u]^1].flow -= a[t];
			u = edges[p[u]].from;
		}
		return 1;
	}
	
	int Mincost(int s, int t, int &flow, int &cost)
	{
		while(spfa(s, t, flow, cost));
		return cost;
	}
};

void readint(int &x)
{
    char c;
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c))
    {
        x = x*10 + c-'0';
        c = getchar();
    }
}

void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

///////////////////////////////////////

MCMF solver;
int n, m, s, t;

int C[maxn];

char str[110][110];

int main()
{
	
	for(;;)
	{
		readint(n);
		if(!n) break;
		for(int i = 1; i <= n; i++) scanf("%d", &C[i]);
		
		solver.init(2*n+10);
		s = 2*n+1, t = 2*n+2;
		
		for(int i = 1; i <= n; i++) scanf("%s", str[i]+1);
		
		for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			if(str[i][j] == '1') solver.AddEdge(i, j+n, 1, -(C[i]^C[j]));
		}
		for(int i = 1; i <= n; i++)
		{
			solver.AddEdge(s, i, 1, 0);
			solver.AddEdge(i+n, t, 1, 0);
			solver.AddEdge(i, t, 1, 0);
		}

		int cost = 0, flow = 0;
		solver.Mincost(s, t, flow, cost);
		printf("%d\n", -cost);
	}
}