「网络流24题」方格取数问题-题解

题目传送门: 「Luogu P2774」方格取数问题

题目大意

$m$行$n$列的方格图,每个方格中有一个正整数
从方格中取数,任意两个数所在方格没有公共边
求取出的数的最大总和

题解

先选择所有方格,然后考虑删去一些方格
相邻$->$奇偶性不同,构成一个二分图,含有两个点集

  1. 从 源点 向 点集$A$ 接一条 容量为点权 的边
  2. 从 点集$B$ 向 汇点 接一条 容量为点权 的边
  3. 从 点集$A$中每个点 向 与其相邻的在点集$B$中的点 接一条 容量为$inf$ 的边(保证不被割)

求出最小割即最大流即可

代码

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#include <bits/stdc++.h>
using namespace std;

inline int read() {
int x = 0; int f = 1; char ch = getchar();
while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
while (isdigit(ch)) {x = x * 10 + ch - 48; ch = getchar();}
return x * f;
}

const int maxn = 10010;
const int inf = 0x3f3f3f3f;
const int go[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int n, m, s, t, d[maxn], cur[maxn], tot;

struct Edge {
int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn];
void add(int from, int to, int cap) {
edges.push_back((Edge) {from, to, cap, 0});
edges.push_back((Edge) {to, from, 0, 0});
int mm = edges.size();
G[from].push_back(mm - 2);
G[to].push_back(mm - 1);
}

bool vis[maxn];
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = 1;
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 (!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;
for (int& i = cur[x]; i < G[x].size(); ++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 dinic(int s, int t) {
int flow = 0;
while (BFS()) {
memset(cur, 0, sizeof(cur));
flow += DFS(s, inf);
}
return flow;
}

int point(int x, int y) {
return (x - 1) * n + y;
}

int main() {
m = read(); n = read();
s = 0; t = n * m + 1;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
int a = read(); tot += a;
if ((i + j) % 2 == 0) {
add(s, point(i, j), a);
for (int k = 0; k < 4; ++k) {
int nx = i + go[k][0];
int ny = j + go[k][1];
if (1 <= nx && nx <= m && 1 <= ny && ny <= n) {
add(point(i, j), point(nx, ny), inf);
}
}
} else {
add(point(i, j), t, a);
}
}
}
int maxflow = dinic(s, t);
printf("%d\n", tot - maxflow);
return 0;
}

「网络流24题」方格取数问题-题解

https://blog.tonycrane.cc/p/40e04941.html

作者

TonyCrane

发布于

2020-04-16

更新于

2020-05-05

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