Harmonious Army

Harmonious Army

时间限制: 1 Sec  内存限制: 128 MB

题目描述

Now, Bob is playing an interesting game in which he is a general of a harmonious army. There are n soldiers in this army. Each soldier should be in one of the two occupations, Mage or Warrior. There are m pairs of soldiers having combination ability. There are three kinds of combination ability. If the two soldiers in a pair are both Warriors, the army power would be increased by a. If the two soldiers in a pair are both Mages, the army power would be increased by c. Otherwise the army power would be increased by b, and b=a/4+c/3, guaranteed that 4|a and 3|c. Your task is to output the maximum power Bob can increase by arranging the soldiers' occupations.

Note that the symbol a|b means that a divides b, e.g. , 3|12 and 8|24.

输入

There are multiple test cases.
Each case starts with a line containing two positive integers n(n≤500) and m(m≤104).
In the following m lines, each line contains five positive integers u,v,a,b,c (1≤u,v≤n,u≠v,1≤a,c≤4×106,b=a/4+c/3), denoting soldiers u and v have combination ability, guaranteed that the pair (u,v) would not appear more than once.
It is guaranteed that the sum of n in all test cases is no larger than 5×103, and the sum of m in all test cases is no larger than 5×104.

输出

For each test case, output one line containing the maximum power Bob can increase by arranging the soldiers' occupations.

样例输入

3 2
1 2 8 3 3
2 3 4 3 6

样例输出

12

---

题意：给n个士兵安排两种职业，有m个关系，关系的贡献与该关系两边的士兵的职业有关，求最大贡献和。
最小割建图好题，Mark一下。思路参考题解：

对每个士兵建立一个点x ，点x 向源点s 连一条边，向汇点t 连一条边，分别表示选择两种职业，然后就可以先加上所有的贡献，通过两点关系用最小割建模，如下图所示。

设一条边的三种贡献为A;B;C，可以得到以下方程：
a+b=A+B(x,y都选Mage)

c+d=C+B(x,y都选Warrior)

a+d+e=A+C(x选Mage,y选Warrior)

b+c+e=A+C(x选Warrior,y选Mage)

可得一组解a=b=(A+B)/2,c=d=(C+B)/2,e=-B+(A+C)/2，然后将所有有关系的两点的图合并，用所有贡献减掉这个图的最小割即可。

#include<bits/stdc++.h>
#define N 505
using namespace std;

typedef struct
{
    int to,next;
    long long flow;
}ss;

ss edg[N*N];
int now_edge=0,s,t;
int head[N];

void addedge(int u,int v,long long flow)
{
    edg[now_edge]=(ss){v,head[u],flow};
    head[u]=now_edge++;

    edg[now_edge]=(ss){u,head[v],0};
    head[v]=now_edge++;
}

int dis[N];

bool bfs()
{
    memset(dis,0,sizeof(dis));
    queue<int>q;
    q.push(s);
    dis[s]=1;

    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=edg[i].next)
        {
            ss &e=edg[i];
            if(e.flow>0&&dis[e.to]==0)
            {
                dis[e.to]=dis[now]+1;
                q.push(e.to);
            }
        }
    }

    if(dis[t]==0)return 0;
    return 1;
}

int current[N];
long long dfs(int x,long long maxflow)
{
    if(x==t)return maxflow;
//    printf("%d %lld\n",x,maxflow);
    for(int i=current[x];i!=-1;i=edg[i].next)
    {
        current[x]=i;
        ss &e=edg[i];
        if(e.flow>0&&dis[e.to]==dis[x]+1)
        {
            long long flow=dfs(e.to,min(maxflow,e.flow));
            if(flow!=0)
            {
                e.flow-=flow;
                edg[i^1].flow+=flow;
                return flow;
            }
        }
    }
    return 0;
}

long long dinic()
{
    long long ans=0,flow;
    while(bfs())
    {
        for(int i=0;i<N;i++)current[i]=head[i];
        while(flow=dfs(s,LLONG_MAX/2))ans+=flow;
    }
    return ans;
}

void init()
{
    for(int i=0;i<N;i++)head[i]=-1;
    now_edge=0;
}

int Map[N][N];

int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)==2)
    {
        init();
        for(int i=0;i<=n+2;i++)
            for(int j=0;j<=n+2;j++)Map[i][j]=0;

        long long ans=0;

        s=n+1;
        t=s+1;

        while(m--)
        {
            int u,v,a,b,c;
            scanf("%d %d %d %d %d",&u,&v,&a,&b,&c);
            ans+=a+b+c;
            Map[s][u]+=a+b;
            Map[s][v]+=a+b;
            Map[u][t]+=c+b;
            Map[v][t]+=c+b;
            Map[u][v]+=a+c-2*b;
            Map[v][u]+=a+c-2*b;
        }

        for(int i=1;i<=n+2;i++)
            for(int j=1;j<=n+2;j++)if(Map[i][j])addedge(i,j,Map[i][j]);

        printf("%lld\n",ans-dinic()/2);
    }
    return 0;
}