BZOJ 3996: [TJOI2015]线性代数

题意：给出一个N*N的矩阵B和一个1*N的矩阵C。求出一个1*N的01矩阵A.使得
D=（A*B-C）*A^T最大。其中A^T为A的转置。输出D

题目描述似乎很厉害…但A矩阵是01矩阵，也就是取和不取，可以从这方面入手。把D的式子化简D=A*B*A^T-C*A^T.
D= (i=1-n) (j=1-n) ai*aj*bij - (i=1-n) ai*ci
ai和aj存在二元关系，可以最小割解决。由S向i连bij/2,j向T连bij/2, i,j互相连bij/2,且S向每个i连Ci.
Tips:避免实数运算，先将答案乘2，再将结果除以2

#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<queue>using namespace std;const int maxn=1000+10;const int INF=1000000000;int ans,n,s,t,m,cur[maxn],dis[maxn],vis[maxn];struct edge{  int from,to,cap,flow;};vector<int> g[maxn];vector<edge> edges;void addedge(int from,int to,int cap){  edges.push_back((edge){from,to,cap,0});  edges.push_back((edge){to,from,0,0});  int m1=edges.size();  g[from].push_back(m1-2);  g[to].push_back(m1-1);}bool bfs(){  memset(vis,0,sizeof(vis));  queue<int> Q;Q.push(s);vis[s]=1;dis[s]=0;  while(!Q.empty())  {    int x=Q.front();Q.pop();    for(int i=0;i<g[x].size();i++)    {      edge e=edges[g[x][i]];      if(e.cap>e.flow&&!vis[e.to])      {        vis[e.to]=1;        dis[e.to]=dis[x]+1;        Q.push(e.to);      }    }  }  return vis[t];}int dfs(int x,int a){  if(x==t||a==0) return a;  int flow=0,f;  for(int &i=cur[x];i<g[x].size();i++)  {    edge &e=edges[g[x][i]];    if(dis[e.to]==dis[x]+1&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0)    {      flow+=f;      e.flow+=f;      edges[g[x][i]^1].flow-=f;      a-=f;      if(a==0) break;    }  }  return flow;}int maxflow(){  int flow=0;  while(bfs())  {    memset(cur,0,sizeof(cur));    flow+=dfs(s,INF);  }  return flow;}int main(){  //freopen("3996.in","r",stdin);  //freopen("3996.out","w",stdout);  scanf("%d",&n);s=0,t=n+1;  for(int i=1;i<=n;i++)    for(int j=1;j<=n;j++)    {      int x;scanf("%d",&x);ans+=2*x;      if(i==j) addedge(s,i,2*x);      else      {        addedge(s,i,x);        addedge(s,j,x);        addedge(i,j,x);        addedge(j,i,x);      }    }  for(int i=1;i<=n;i++)  {    int x;scanf("%d",&x);    addedge(i,t,2*x);  }  printf("%d\n",(ans-maxflow())/2);  return 0;}