ACdream-1171 Matrix sum， 最大费用最大流

参考:http://blog.csdn.net/yew1eb/article/details/38360253
题目链接:
http://acdream.info/problem?pid=1171

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。

心得:注意这个地方的最大费用最大流要掌握原理:

#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;}