UVA - 11082 Matrix Decompressing （最大流，技巧）

很经典的网络流模型，行编号和列编号分别看成一个点，行和列和分别看出容量，一个点(x,y)看出是一条边，边的容量下界是1，所以先减去1，之后在加上就好了。

建图的时候注意分配好编号，解从残留网络中的边找。

前向星建图的话，打印解会比较麻烦。

#include<bits/stdc++.h>
using namespace std;

const int maxn = 42;

struct Edge
{
int v,cap,flow;
};

vector<Edge> edges;
#define PB push_back

vector<int> G[maxn];

void AddEdge(int u,int v,int c)
{
G[u].PB(edges.size());
edges.PB(Edge{v,c,0});
G[v].PB(edges.size());
edges.PB(Edge{u,0,0});
}

const int INF = 0x3f3f3f3f;
int S,T;
int vcnt;
bool vis[maxn];
int d[maxn];
int q[maxn<<1];

bool bfs()
{
memset(vis,0,sizeof(bool)*vcnt);
int l = 0,r = 0;
q[r++] = S; vis[S] = true;
while(r>l){
int u = q[l++];
for(int i = 0; i < G[u].size(); i++){
Edge &e = edges[G[u][i]];
if(!vis[e.v] && e.cap >e.flow){
vis[e.v] = true;
d[e.v] = d[u]+1;
q[r++] = e.v;
}
}
}
return vis[T];
}

int cur[maxn];
int dfs(int u,int a)
{
if(u == T||!a) return a;
int flow = 0, f;
for(int &i = cur[u]; i < G[u].size(); i++){
Edge &e = edges[G[u][i]];
if(d[e.v] == d[u]+1 && (f = dfs(e.v,min(a,e.cap-e.flow)))>0){
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(!a) break;
}
}
return flow;
}

int MaxFlow()
{
int flow = 0;
while(bfs()){
memset(cur,0,sizeof(int)*vcnt);
flow += dfs(S,INF);
}
return flow;
}

const int N = 20;
int rid
,cid
;

void init()
{
vcnt = 2;
edges.clear();
}

int main()
{
//freopen("in.txt","r",stdin);
S = 0; T = 1;
int testCase; scanf("%d",&testCase);
int mcnt = 0;
while(testCase--){
init();
int R,C,cap,pre; scanf("%d%d",&R,&C);

for(int i = 0, M = R+C+2; i < M; i++) G[i].clear();
pre = 0;
for(int i = 0; i < R; i++) {
rid[i] = vcnt++;
scanf("%d",&cap);
AddEdge(S,rid[i],cap-pre-C);
pre = cap;
}
pre = 0;
for(int i = 0; i < C; i++){
cid[i] = vcnt++;
scanf("%d",&cap);
AddEdge(cid[i],T,cap-pre-R);
pre = cap;
}

for(int i = 0; i < R; i++)
for(int j = 0; j < C; j++){
AddEdge(rid[i],cid[j],19);
}

MaxFlow();

printf("Matrix %d\n",++mcnt);
for(int i = 0; i < R; i++){
int k = 0, u = rid[i];
for(int j = 0; j < G[u].size(); j++){
Edge &e = edges[G[u][j]];
if(e.v == cid[k]){
printf("%d%c",e.flow+1,++k==C?'\n':' ');
if(k == C) break;
}
}
}

if(testCase) putchar('\n');
}
return 0;
}