题目大意:给你一个n*n的格子的棋盘,每个格子里面有一个非负数。
从中取出若干个数,使得任意的两个数所在的格子没有公共边,就是说所取的数所在的2个格子不能相邻,并且取出的数的和最大。
解题思路:最大点权独立集,关键是怎么建图了,我们可以采用染色的思想对这张图进行染色,然后分成两个点集
假设将第一个格子染成白色,然后将它相邻的格子染成相反的颜色黑色,以此类推,这样就可以将一张图分成染成黑白两种颜色的点集了
然后就是连边了,连边的话,我们只考虑白色格子的连向黑色格子的,因为两者之间是相对的,所以只需要取一条就好了
这样图就建好了
最大权值就是:总权值-最小割了(最小割就是最小点权覆盖了)具体的证明请看胡伯涛:算法合集之《最小割模型在信息学竞赛中å的应用》
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define M 1000010
#define N 10010
#define INF 0x3f3f3f3f
struct Edge{
int u, v, cap, flow, next;
Edge() {}
Edge(int u, int v, int cap, int flow, int next): u(u), v(v), cap(cap), flow(flow), next(next) {}
}E[M];
struct Dinic{
int head[N], d[N];
int tot, sink, source;
void init() {
memset(head, -1, sizeof(head));
tot = 0;
}
inline void AddEdge(int u, int v, int cap) {
E[tot] = Edge(u, v, cap, 0, head[u]); head[u] = tot++;
u = u ^ v; v = u ^ v; u = u ^ v;
E[tot] = Edge(u, v, 0, 0, head[u]); head[u] = tot++;
}
inline bool bfs(int s) {
int u, v;
memset(d, 0, sizeof(d));
queue<int> Q;
Q.push(s);
d[s] = 1;
while (!Q.empty()) {
u = Q.front(); Q.pop();
if (u == sink) return true;
for (int i = head[u]; ~i; i = E[i].next) {
v = E[i].v;
if (!d[v] && E[i].cap - E[i].flow > 0) {
d[v] = d[u] + 1;
Q.push(v);
}
}
}
return false;
}
int dfs(int x, int a) {
if (x == sink || a == 0)
return a;
int f, flow = 0;
for (int i = head[x]; ~i; i = E[i].next) {
int v = E[i].v;
if (d[v] == d[x] + 1 && E[i].cap - E[i].flow > 0) {
f = dfs(v, min(a, E[i].cap - E[i].flow));
E[i].flow += f;
E[i^1].flow -= f;
flow += f;
a -= f;
if (!a) break;
}
}
if (flow == 0) d[x] = 0;
return flow;
}
int Maxflow(int source, int sink) {
int flow = 0;
this->sink = sink;
while (bfs(source)) flow += dfs(source, INF);
return flow;
}
};
Dinic dinic;
#define maxn 25
int n, source, sink, Sum;
int num[maxn][maxn];
bool mark[maxn][maxn];
int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
void init() {
source = 0, sink = n * n + 1, Sum = 0;
dinic.init();
for (int i = 0; i < n; i++)
for (int j = 1; j <= n; j++) {
scanf("%d", &num[i][j]);
Sum += num[i][j];
}
memset(mark, 0, sizeof(mark));
for (int i = 0; i < n; i++)
for (int j = 1; j <= n; j++)
if (!mark[i][j]) {
dinic.AddEdge(source, i * n + j, num[i][j]);
for (int k = 0; k < 4; k++) {
int x = i + dir[k][0], y = j + dir[k][1];
if (x < 0 || x >= n || y < 1 || y > n) continue;
dinic.AddEdge(i * n + j, x * n + y, INF);
if (!mark[x][y]) dinic.AddEdge(x * n + y, sink, num[x][y]);
mark[x][y] = true;
}
}
printf("%d\n", Sum - dinic.Maxflow(source, sink));
}
int main() {
while (scanf("%d", &n) != EOF) {
init();
}
return 0;
}