【BZOJ3144】[Hnoi2013]切糕【最小割】

来源：互联网发布：汉字笔顺在线演示软件编辑：程序博客网时间：2024/04/24 23:51

【题目链接】

学习一发建图。

这篇题解比较详细【zarxdy34的题解】

/* Telekinetic Forest Guard */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxg = 45, maxn = 70005, maxm = 150005, maxq = 100000, inf = 0x3f3f3f3f;int n, m, h, d, head[maxn], cur[maxn], cnt, bg, ed, depth[maxn], q[maxq], id[maxg][maxg][maxg];int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};struct _edge {int v, w, next;} g[maxm << 1];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void add(int u, int v, int w) {g[cnt] = (_edge){v, w, head[u]};head[u] = cnt++;}inline void insert(int u, int v, int w) {add(u, v, w); add(v, u, 0);}inline bool bfs() {for(int i = 0; i <= ed; i++) depth[i] = -1;int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {depth[g[i].v] = depth[u] + 1;if(g[i].v == ed) return 1;q[t++] = g[i].v;}return 0;}inline int dfs(int x, int flow) {if(x == ed) return flow;int left = flow;for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {int tmp = dfs(g[i].v, min(left, g[i].w));left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;if(g[i].w) cur[x] = i;if(!left) return flow;}if(left == flow) depth[x] = -1;return flow - left;}int main() {n = iread(); m = iread(); h = iread(); d = iread();for(int i = 0; i < maxn; i++) head[i] = -1; cnt = 0;int tot = 0; bg = ++tot;for(int i = 1; i <= h + 1; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= m; k++)id[i][j][k] = ++tot;ed = ++tot;for(int i = 1; i <= h; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= m; k++) {intw = iread();insert(id[i][j][k], id[i + 1][j][k], w);if(i + d <= h + 1) for(int r = 0; r < 4; r++) {int x = j + dx[r], y = k + dy[r];if(x < 1 || x > n || y < 1 || y > m) continue;insert(id[i + d][x][y], id[i][j][k], inf);}}for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) insert(bg, id[1][i][j], inf), insert(id[h + 1][i][j], ed, inf);int ans = 0;while(bfs()) {for(int i = 0; i <= ed; i++) cur[i] = head[i];ans += dfs(bg, inf);}printf("%d\n", ans);return 0;}

0 0