题目链接：点击打开链接
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)
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
6
n行作为左端点 m作为右端点建一个二部图
n个点连到源点 流量为inf ，费用为0
m个点连到汇点 流量为inf 费用为0
n个点和m个点之间连一条费用为 mp[i][j] ，流量为1的点
--------------------------- 以上是普通建图-----------------------
为了达到有下界的效果
给下界对应的点加一个费用为-inf，流量为ki的边，这样让费用流强制把所有费用为-inf的边先跑
意思也就是先使得下界的边满流。
当费用>0时， 说明这次跑的边中不存在有下界的边，那么就相当于下界的边已经满流，所以直接终止费用流
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
#include<vector>
#define eps 1e-9
#define pi acos(-1.0)
using namespace std;
#define ll int
#define inf 0x3f3f3f3f
#define Inf 0x3FFFFFFFFFFFFFFFLL
#define N 105
#define M 1005
struct Edge
{
ll from,to,cap,cost,nex;
Edge(){}
Edge(ll from,ll to,ll cap,ll cost,ll next):from(from),to(to),cap(cap),cost(cost),nex(next){}
}edges[M<<1];
ll head[N], edgenum;
ll d[N], a[N], p[N];
bool inq[N];
void add(ll from,ll to,ll cap,ll cost)
{
edges[edgenum] = Edge(from,to,cap,cost,head[from]);
head[from] = edgenum++;
edges[edgenum] = Edge(to,from,0,-cost,head[to]);
head[to] = edgenum++;
}
bool spfa(ll s, ll t, ll &flow, ll &cost)
{
for(ll i = 0; i <= t; i++) d[i] = inf;
memset(inq, 0, sizeof inq);
queue<ll>q;
q.push(s);
d[s] = 0; a[s] = inf;
while(!q.empty())
{
ll u = q.front(); q.pop();
inq[u] = 0;
for(ll i = head[u]; ~i; i = edges[i].nex)
{
Edge &e = edges[i];
if(e.cap && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = i;
a[e.to] = min(a[u], e.cap);
if(!inq[e.to])
{inq[e.to]=1; q.push(e.to);}
}
}
}
if(d[t]>0) return false;
cost += d[t] * a[t];
flow += a[t];
ll u = t;
while(u != s){
edges[ p[u] ].cap -= a[t];
edges[p[u]^1].cap += a[t];
u = edges[p[u]^1].to;
}
return true;
}
ll Mincost(ll s,ll t){//返回最小费用
ll flow = 0, cost = 0;
while(spfa(s, t, flow, cost));
return cost;
}
void init(){memset(head,-1,sizeof head); edgenum = 0;}
int n,m;
int mp[55][55];
int h[55], l[55];
void input(){
scanf("%d %d",&n,&m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
scanf("%d",&mp[i][j]);
for(int i = 1; i <= n; i++)scanf("%d",&h[i]);
for(int i = 1; i <= m; i++)scanf("%d",&l[i]);
}
#define hehe 100000
int main(){
int T, i, j;scanf("%d",&T);
while(T--){
input();
init();
int from = 0, to = n+m+1;
int les = 0;
for(i = 1; i <= n; i++)
{
les += h[i];
add(from, i, h[i], -hehe);
add(from, i, inf, 0);
}
for(i = 1; i <= m; i++)
{
les += l[i];
add(n + i, to, l[i], -hehe);
add(n + i, to, inf, 0);
}
for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
add(i, n+j, 1, mp[i][j]);
printf("%d\n", Mincost(from, to) + hehe*les);
}
return 0;
}
/*
http://acdream.info/onecontest/1080#problem-H
http://paste.ubuntu.com/7930356/#userconsent#
*/Acdream 1171 Matrix sum 上下界费用流,布布扣,bubuko.com
Acdream 1171 Matrix sum 上下界费用流
原文：http://blog.csdn.net/qq574857122/article/details/38353251