【loj】#6007. 「网络流 24 题」方格取数(二分图最大点权独立集)

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记录一个菜逼的成长。。

题目链接

这里写图片描述

算法合集之《最小割模型在信息学竞赛中的应用》
在上面的论文末尾里有对这个问题的具体解释

#include <bits/stdc++.h>using namespace std;#define rep(i,l,r) for( int i = l; i <= r; i++ )#define rep0(i,l,r) for( int i = l; i < r; i++ )#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define clr clear()#define pb push_back#define mp make_pair#define fi first#define se secondtypedef long long LL;typedef pair<int,int> PII;const int INF = 0x3f3f3f3f;const int MAX_V = 1000 + 10;struct edge{  int to,cap,rev,flow;  edge(){}  edge(int _to,int _cap,int _rev,int _flow):to(_to),cap(_cap),rev(_rev),flow(_flow){}};vector<edge>G[MAX_V];int level[MAX_V];int iter[MAX_V];void add(int from,int to,int cap,int flow = 0){  G[from].push_back(edge(to,cap,G[to].size(),0));  G[to].push_back(edge(from,0,G[from].size()-1,0));}void bfs(int s){  memset(level,-1,sizeof(level));  queue<int>que;  level[s] = 0;  que.push(s);  while(!que.empty()){    int f = que.front();    que.pop();    for( int i = 0; i < G[f].size(); i++ ){      edge &e = G[f][i];      if(e.cap > 0 && level[e.to] == -1){        level[e.to] = level[f] + 1;        que.push(e.to);      }    }  }}int dfs(int v,int t,int f){  if(v == t)return f;  for( int &i = iter[v]; i < G[v].size(); i++ ){    edge &e = G[v][i];    if(e.cap > 0 && level[v] < level[e.to]){      int d = dfs(e.to,t,min(e.cap,f));      if(d > 0){        e.cap -= d;e.flow += d;        G[e.to][e.rev].cap += d;        return d;      }    }  }  return 0;}int max_flow(int s,int t){  int flow = 0;  for(;;){    bfs(s);    if(level[t] == -1)return flow;    memset(iter,0,sizeof(iter));    int f;    while((f = dfs(s,t,INF)) > 0)      flow += f;  }}int a[MAX_V][MAX_V];int n,m;int getID(int i,int j){  return (i-1) * m + j;}int main(){  scanf("%d%d",&n,&m);  int sum = 0,s = 0,t = n * m + 1;  rep(i,1,n){    rep(j,1,m){      int x;      scanf("%d",&x);      sum += x;      int flag = !((i+j) & 1);      int id = getID(i,j);      if(flag)add(s,id,x);      else add(id,t,x);      if(flag){        if(i > 1)add(id,getID(i-1,j),INF);        if(i < n)add(id,getID(i+1,j),INF);        if(j > 1)add(id,getID(i,j-1),INF);        if(j < m)add(id,getID(i,j+1),INF);      }    }  }  printf("%d\n",sum - max_flow(s,t));  return 0;}

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