HDU 1569 方格取数 (最小割)

Posted 2020-10-09 dwtfukgv

题意：中文题。

析：很明显的是二分图的最大独立集，但是每个点都有权值，这个可以用最小割来求，建立一个超级源点s，和汇点t，然后s 向 X集，添加容量为权值的边，Y集向 t 添加容量为权值的，然后跑一遍最小割，然后用总权值减去就是答案了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50 * 50 + 20;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int from, to, cap, flow;
};

struct Dinic{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];

  void init(int n){
    this-> n = n;
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int cap){
    edges.pb((Edge){from, to, cap, 0});
    edges.pb((Edge){to, from, 0, 0});
    m = edges.sz;
    G[from].pb(m-2);
    G[to].pb(m-1);
  }

  bool bfs(){
    queue<int> q;
    ms(vis, 0);  d[s] = 0;
    q.push(s);  vis[s] = 1;
    while(!q.empty()){
      int x = q.front();  q.pop();
      for(int i = 0; i < G[x].sz; ++i){
        Edge &e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[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 = 0;
    for(int &i = cur[x]; i < G[x].sz; ++i){
      Edge &e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0)  break;
      }
    }
    return flow;
  }

  int maxflow(int s, int t){
    this->s = s;
    this->t = t;
    int flow = 0;
    while(bfs()){ ms(cur, 0);  flow += dfs(s, INF);  }
    return flow;
  }
};
Dinic dinic;

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    dinic.init(n * m + 10);
    int s = 0, t = n * m + 5;
    int sum = 0;
    FOR(i, 0, n)  for(int j = 1; j <= m; ++j){
      int x;
      scanf("%d", &x);
      sum += x;
      int now = i * m + j;
      if(i + j & 1){
        dinic.addEdge(s, now, x);
        if(i)  dinic.addEdge(now, now - m, INF);  // up
        if(j > 1)  dinic.addEdge(now, now - 1, INF);  // left
        if(i + 1 < n)  dinic.addEdge(now, now + m, INF); // down
        if(j < m)  dinic.addEdge(now, now + 1, INF); // right
      }
      else dinic.addEdge(now, t, x);
    }
    printf("%d\n", sum - dinic.maxflow(s, t));
  }
  return 0;
}