版权声明:https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/88739482
题目链接
另外一个提交链接
这道题要是不看别的博主的说明的话,根本就想不到,用到了所谓的点覆盖模型,点覆盖模型的略微解释:两相邻点只能取其中一点,存在着这样的限制的取点方法就是了。
那么,我们该如何去解?看了大牛的博客,然后模拟了一下,才知道大致应该怎么做,当然,思维的确好好,我们将点分为黑白点,其中,黑点的周围都是白点,同样的,白点的周围的点都是黑点,我们利用这样的方式来解这个问题我们从源点链接到黑点边容量为权值的边,在又白点链接到汇点,为边容量为点权值的边,然后对于每个黑点,链接上每个白点,边权为INF,这样子,我们只需要去求一个最小割即可,我们利用全体点权之和再减去最小割的量得到的就是最后我们需要的最大值了。
这种求最大值且有限制的问题一般都需要转移成总和-最小割的形式(这几道题的反馈得到的思维)。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxE = 6e4 + 7, maxN = 1e4 + 7, S = 0;
const int dir[4][2] =
{
-1, 0,
0, -1,
0, 1,
1, 0
};
int head[maxN], cur[maxN], cnt, N, M, T;
struct Eddge
{
int nex, to, flow;
Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c) {}
}edge[maxE<<1];
inline void addEddge(int u, int v, int val)
{
edge[cnt] = Eddge(head[u], v, val);
head[u] = cnt++;
}
bool In_Map(int x, int y) { return x > 0 && x <= N && y > 0 && y <= M; }
inline void _ADD(int x, int y)
{
int xx, yy, now = M * (x - 1) + y, to;
for(int i=0; i<4; i++)
{
xx = x + dir[i][0]; yy = y + dir[i][1];
to = M * (xx - 1) + yy;
if(In_Map(xx, yy))
{
addEddge(now, to, INF);
addEddge(to, now, 0);
}
}
}
int deep[maxN];
queue<int> Q;
bool bfs()
{
while(!Q.empty()) Q.pop();
memset(deep, 0, sizeof(deep)); deep[S] = 1;
Q.push(S);
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i=head[u], v, flow; ~i; i=edge[i].nex)
{
v = edge[i].to; flow = edge[i].flow;
if(!deep[v] && flow)
{
deep[v] = deep[u] + 1;
Q.push(v);
}
}
}
return deep[T];
}
int dfs(int u, int flow)
{
if(u == T) return flow;
for(int &i=cur[u], v, val, dist; ~i; i=edge[i].nex)
{
v = edge[i].to; val = edge[i].flow;
if(deep[v] == deep[u] + 1 && val)
{
dist = dfs(v, min(flow, val));
if(dist)
{
edge[i].flow -= dist;
edge[i^1].flow += dist;
return dist;
}
}
}
return 0;
}
int Dinic()
{
int ans = 0, tmp = 0;
while(bfs())
{
for(int i = S; i <= T; i++) cur[i] = head[i]; //memcpy(cur, head, sizeof(cur));
while((tmp = dfs(S, INF))) ans += tmp;
}
return ans;
}
inline void init()
{
cnt = 0; T = N * M + 1;
memset(head, -1, sizeof(head));
}
int main()
{
scanf("%d%d", &N, &M);
init();
int all = 0;
for(int i=1, e1, tmp; i<=N; i++)
{
for(int j=1; j<=M; j++)
{
scanf("%d", &e1); all += e1;
tmp = M * (i - 1) + j;
if((i ^ j) & 1)
{
addEddge(tmp, T, e1);
addEddge(T, tmp, 0);
}
else
{
addEddge(S, tmp, e1);
addEddge(tmp, S, 0);
_ADD(i, j);
}
}
}
printf("%d\n", all - Dinic());
return 0;
}