ACdream-1171 Matrix sum， 最大费用最大流

Matrix sum

Problem Description

Input

Output

Sample Input

1
4 4
1 1 1 1
1 10 10 10
1 10 10 10
1 10 10 10
1 1 1 1
1 1 1 1

Sample Output

6

分析：

很容易想到二分图模型（n行左端点，m列右端点） --> 有上下界的费用流

每行每列取数的个数不能少于R[i] / C[i]， 问取得数总和最小是多少Min_Sum？ 

转化为

每行每列取数的个数不多于 m-R[i] / n - C[i]，问取得数总和最大是多少Max_Sum？

Min_Sum = All_Sum - Max_Sum 

                     总数 - 最大费用最大流即可

这样就把有上下界的费用流问题转化为（只有上界）普通的费用流问题了。

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn = 202 + 10;
const int INF = 1000000000;

typedef long long LL;

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

struct MCMF { //<span style="font-size:14px;">最大费用最大流</span>
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];         // 是否在队列中
  int d[maxn];           // Bellman-Ford
  int p[maxn];           // 上一条弧
  int a[maxn];           // 可改进量

  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 BellmanFord(int s, int t, LL& ans) {
    for(int i = 0; i <= t; i++) d[i] = -INF;  //与最小费用最大流相反（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) { <span style="font-family: Arial, Helvetica, sans-serif;">//与最小费用最大流相反（d[e.to] < d[u] + e.cost ）</span>
          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] < 0) return false;  //<span style="font-family: Arial, Helvetica, sans-serif;">与最小费用最大流相反(d[i]>=0)</span>
    ans += (LL)d[t] * (LL)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 true;
  }

  // 需要保证初始网络中没有负权圈
  LL Mincost(int s, int t) {
    LL cost = 0;
    while(BellmanFord(s, t, cost));
    return cost;
  }

};

MCMF g;

int S, T;
LL SUM;
int a[60][60], R[60], C[60];

void init()
{
    int n, m, i, j;
    scanf("%d%d", &n, &m);
    SUM = 0;
    for(i=1; i<=n; ++i)
    for(j=1; j<=m; ++j)
    {
        scanf("%d", &a[i][j]);
        SUM += a[i][j];
    }
    for(i=1; i<=n; ++i) scanf("%d", &R[i]);
    for(i=1; i<=m; ++i) scanf("%d", &C[i]);

    S = 0, T = n + m + 1;
    g.init(T);
    for(i=1; i<=n; ++i)
    {
        g.AddEdge(S, i, m-R[i], 0);
    }
    for(i=1; i<=m; ++i)
    {
        g.AddEdge(i+n, T, n-C[i], 0);
    }

    for(i=1; i<=n; ++i)
    for(j=1; j<=m; ++j)
    {
        g.AddEdge(i, j+n, 1, a[i][j]);
    }
}

void solve()
{
    LL t = g.Mincost(S, T);
    printf("%I64d\n", SUM - t);
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        init();
        solve();
    }
    return 0;
}

官方题解   最大费用最大流板子

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int E = 50010;
const int oo = 0x7fffffff;
const int N = 210;

struct edge
{
    int next,v,flow,cost;
}e[E];

int head[N],cnt;
queue<int> q;

void addedge(int u,int v,int flow,int cost)
{
    e[cnt].v = v;
    e[cnt].flow = flow;
    e[cnt].cost = cost;
    e[cnt].next = head[u];
    head[u] = cnt ++;
}

void addEdge(int u,int v,int flow,int cost)
{
    addedge(u,v,flow,cost);
    addedge(v,u,0, -cost);
}

int S,T;
int ans;
int a[N][N];

void init()
{
    int n,m;
    scanf("%d%d",&n,&m);
    ans = 0;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++) {
            scanf("%d",&a[i][j]);
            ans += a[i][j];
        }
    int R[N],C[N];
    for(int i = 1; i <= n; i ++) scanf("%d",&R[i]);
    for(int i = 1; i <= m; i ++) scanf("%d",&C[i]);
    S = 0,T = n + m + 1;
    cnt = 0;
    memset(head,-1,sizeof(head));
    for(int i = 1; i <= n; i ++) {
        addEdge(S,i,m - R[i],0);
    }
    for(int i = 1; i <= m; i ++)
        addEdge(i + n,T,n - C[i],0);
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            addEdge(i,j + n,1,a[i][j]);
}

int dis[N],cc[N],visit[N],pre[N],dd[N];

int spfa()
{
    fill(dis,dis + T + 1, -oo);
    dis[S] = 0;
    pre[S] = -1;
    while(!q.empty()) q.pop();
    q.push(S);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        visit[u] = 0;
        for(int i = head[u]; i != -1; i = e[i].next) {
            if(e[i].flow > 0 && dis[e[i].v] < dis[u] + e[i].cost) {
                dis[e[i].v] = dis[u] + e[i].cost;
                pre[e[i].v] = u;
                cc[e[i].v] = i;
                dd[e[i].v] = e[i].cost;
                if(!visit[e[i].v]) {
                    q.push(e[i].v);
                    visit[e[i].v] = 1;
                }
            }
        }
    }
    return dis[T] >= 0;
}

int argument()
{
    int aug = oo;
    int u,v;
    int ans = 0;
    for(u = pre[v = T]; v != S; v = u, u = pre[v])
        if(e[cc[v]].flow < aug) aug = e[cc[v]].flow;
    for(u = pre[v = T]; v != S; v = u, u = pre[v]) {
        e[cc[v]].flow -= aug;
        e[cc[v] ^ 1].flow += aug;
        ans += dd[v] * aug;
    }
    return ans;
}

void mcmf()
{
    memset(visit,0,sizeof(visit));
    while(spfa()) {
        int cost = argument();
        if(ans < 0) break;
        ans -= cost;
    }
    printf("%d\n",ans);
}

int main()
{
    //freopen("choose.in","r",stdin);
    //freopen("choose.out","w",stdout);
    int t;
    scanf("%d",&t);
    for(int cas = 1; cas <= t; cas ++) {
        init();
        mcmf();
    }
    return 0;
}

ACdream-1171 Matrix sum， 最大费用最大流,布布扣,bubuko.com

ACdream-1171 Matrix sum， 最大费用最大流

原文：http://blog.csdn.net/yew1eb/article/details/38360253