HDU 3395 Special Fish 最“大”费用最大流

求最大费用能够将边权取负以转化成求最小费用。

然而此时依旧不正确。由于会优先寻找最大流。可是答案并不一定出如今满流的时候。所以要加一些边（下图中的红边）使其在答案出现时满流。

设全部边的流量为1，花费例如以下图所看到的。

显然最大花费是1001。而没有红边的情况下会得到3。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 6000007

using namespace std;

struct E
{
    int u,v,Max,cost,next;
}edge[200100];

int head[310];
int cur[310];

int Top;

void Link(int u,int v,int w,int cost)
{
    edge[Top].u = u;
    edge[Top].v = v;
    edge[Top].cost = cost;
    edge[Top].Max = w;
    edge[Top].next = head[u];
    head[u] = Top++;
}

struct Q
{
    int v;
//    bool operator < (const Q &a) const {
//        return a.cost < cost;
   // }
};

void Updata(int site,int flow,int &cost)
{
    for(int i = site;cur[i] != -1; i = edge[cur[i]^1].v)
    {
        edge[cur[i]].Max -= flow;
        edge[cur[i]^1].Max += flow;
        cost += edge[cur[i]].cost*flow;
    }
}

int dis[310];
int flow[310];

bool mark[310];

queue<Q> q;

int Spfa(int S,int T,int n,int &cost)
{
    while(q.empty() == false)
        q.pop();

    memset(mark,false,sizeof(bool)*(n+2));
    memset(dis,INF,sizeof(int)*(n+2));
    memset(flow,INF,sizeof(int)*(n+2));
    dis[S] = 0;

    Q f,t;

    f.v = S;
    dis[S] = 0;
    cur[S] = -1;
    mark[S] = true;
    q.push(f);

    while(q.empty() == false)
    {
        f = q.front();
        q.pop();
        mark[f.v] = false;
        for(int p = head[f.v]; p != -1; p = edge[p].next)
        {
            t.v = edge[p].v;

            if(edge[p].Max && dis[t.v] > dis[f.v] + edge[p].cost)
            {
                cur[t.v] = p;
                dis[t.v] = dis[f.v] + edge[p].cost;
                flow[t.v] = min(flow[f.v],edge[p].Max);

                if(mark[t.v] == false)
                {
                    mark[t.v] = true;
                    q.push(t);
                }
            }
        }
    }

    if(dis[T] != INF)
    {
        Updata(T,flow[T],cost);
        return flow[T];
    }

    return 0;
}

int Cal_MaxFlow_MinCost(int S,int T,int n)
{
    int f = 0,cost = 0,temp;

    do
    {
        temp = Spfa(S,T,n,cost);
        f += temp;
    }while(temp);

    printf("%d
",-cost);

    return cost;
}

int val[110];

int main()
{
    int n;

    int u,v,w,i,j,x;

    while(scanf("%d",&n) && n)
    {
        memset(head,-1,sizeof(int)*(2*n+3));
        Top = 0;
        for(i = 1;i <= n; ++i)
        {
            scanf("%d",&val[i]);
        }

        int S = 1,T = 2*n+2;

        for(i = 1;i <= n; ++i)
        {
            Link(S,i+1,1,0);
            Link(i+1,S,0,0);
            Link(n+i+1,T,1,0);
            Link(T,n+i+1,0,0);
            Link(i+1,T,1,0);
            Link(T,i+1,0,0);
        }

        for(i = 1;i <= n; ++i)
        {
            for(j = 1;j <= n; ++j)
            {
                scanf("%1d",&x);
                if(x)
                {
                    Link(i+1,j+n+1,1,-(val[i]^val[j]));
                    Link(j+n+1,i+1,0,(val[i]^val[j]));
                }
            }
        }
        Cal_MaxFlow_MinCost(S,T,T);
    }
    return 0;
}