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)
Submit Statistic Next Problem
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
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;
}