HDU 1565 1569 方格取数（最大点权独立集）
题意：中文题
思路：最大点权独立集 = 总权值 - 最小割 = 总权值 - 最大流
那么原图周围不能连边，那么就能够分成黑白棋盘。源点连向黑点。白点连向汇点，容量都为点容量。然后黑白之间相邻的就连一条容量无限大的边
代码：
#include#include #include #include using namespace std;const int MAXNODE = 2505;const int MAXEDGE = 100005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; }};struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; int cur[MAXNODE]; vector cut; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; Type flow = 0; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } void MinCut() { cut.clear(); for (int i = 0; i < m; i += 2) { if (vis[edges[i].u] && !vis[edges[i].v]) cut.push_back(i); } }} gao;const int N = 55;const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};int n, m, g[N][N], sum;int main() { while (~scanf("%d%d", &n, &m)) { sum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) { scanf("%d", &g[i][j]); sum += g[i][j]; } gao.init(n * m + 2); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if ((i + j) % 2 == 0) { gao.add_Edge(0, i * m + j + 1, g[i][j]); for (int k = 0; k < 4; k++) { int x = i + d[k][0]; int y = j + d[k][1]; if (x < 0 || x >= n || y < 0 || y >= m) continue; gao.add_Edge(i * m + j + 1, x * m + y + 1, g[i][j] + g[x][y]); } } else gao.add_Edge(i * m + j + 1, n * m + 1, g[i][j]); } } printf("%d\n", sum - gao.Maxflow(0, n * m + 1)); } return 0;}