題目鏈接

結論

$|帶權最大獨立集|=|點權之和|?|最小割|=|點權之和|?|最大流|$

思路

黑白染色之后建圖，源點$S$到黑點建對應權值邊，白點到匯點$T$建對應取值邊，黑點到相鄰的白點建$inf$邊，然后求得的最大流，就是將黑白分開的最小割。

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 #define endl '\n' const int maxn = 50 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; struct ac{int v, c, pre; }edge[maxn<<8]; int s, e; int head[maxn<<8], dis[maxn<<6], curedge[maxn<<8], cnt; void init() {mem(head, -1);cnt = 0; } void addedge(int u, int v, int c) { // 記得雙向邊edge[cnt] = {v, c, head[u]};head[u] = cnt++; } bool bfs() {queue<int> que;que.push(s);mem(dis, 0);dis[s] = 1;while (!que.empty()) {int f = que.front();que.pop();for (int i = head[f]; i != -1; i = edge[i].pre) {if (dis[edge[i].v] || edge[i].c == 0) continue;dis[edge[i].v] = dis[f] + 1;que.push(edge[i].v);}}return dis[e] > 0; }int dfs(int now, int flow) {if (now == e || flow == 0) return flow;for (int &i = curedge[now]; i != -1; i = edge[i].pre) { // 當前弧優化if (dis[edge[i].v] != dis[now] + 1 || edge[i].c == 0) continue;int d = dfs(edge[i].v, min(flow, edge[i].c));if (d > 0) {edge[i].c -= d;edge[i^1].c += d;return d;} }dis[now] = -1; // // 炸點優化return 0; } int Dinic() {int sum = 0, d;while (bfs()) {for (int i = 0; i <= e; ++i) curedge[i] = head[i];while (d = dfs(s, inf)) sum += d;}return sum; }int a[maxn][maxn]; int main () {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);int n, m;while (cin >> n >> m) {init();int sum = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {cin >> a[i][j];sum += a[i][j];}}s = 0, e = n * m + 1;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if ((i + j) % 2 == 0) {addedge(s, i*m+j+1, a[i][j]);addedge(i*m+j+1, s, 0);for (int k = -1; k <= 1; ++k) {for (int h = -1; h <= 1; ++h) {if (abs(k) + abs(h) != 1) continue;int x = i + k;int y = j + h;if (x < 0 || y < 0 || x >= n || y >= m) continue;if ((x + y) % 2 == 0) continue;addedge(i*m+j+1, x*m+y+1, inf);addedge(x*m+y+1, i*m+j+1, 0);}}}else {addedge(i*m+j+1, e, a[i][j]);addedge(e, i*m+j+1, 0);} }}int ans = sum - Dinic();cout << ans << endl;}return 0; }

總結

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