ACdream-1171 Matrix sum， 最大费用最大流

Matrix sum

Time Limit: 8000/4000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

sweet和zero在玩矩阵游戏，sweet画了一个N * M的矩阵，矩阵的每个格子有一个整数。zero给出N个数Ki，和M个数Kj，zero要求sweet选出一些数，满足从第 i 行至少选出了Ki个数，第j列至少选出了Kj个数。 这些数之和就是sweet要付给zero的糖果数。sweet想知道他至少要给zero多少个糖果，您能帮他做出一个最优策略吗？

Input

首行一个数T(T <= 40)，代表数据总数，接下来有T组数据。

每组数据：

第一行两个数N,M(1 <= N,M <= 50)

接下来N行，每行M个数（范围是0-10000的整数)

接下来一行有N个数Ki,表示第i行至少选Ki个元素(0 <= Ki <= M)

最后一行有M个数Kj，表示第j列至少选Kj个元素(0 <= Kj <= N)

Output

每组数据输出一行，sweet要付给zero的糖果数最少是多少

Sample Input

14 41 1 1 11 10 10 101 10 10 101 10 10 101 1 1 11 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;}