POJ 2375 Cow Ski Area（强连通）

POJ 2375 Cow Ski Area

id=2375" target="_blank" style="">题目链接

题意：给定一个滑雪场，每一个点能向周围4个点高度小于等于这个点的点滑，如今要建电缆，使得随意两点都有路径互相可达，问最少须要几条电缆

思路：强连通缩点。每一个点就是一个点。能走的建边。缩点后找入度出度为0的个数的最大值就是答案。注意一開始就强连通了答案应该是0

代码：

#include <cstdio>
#include <cstring>
#include <stack>
#include <algorithm>
using namespace std;

const int N = 250005;
const int M = 1000005;
const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};

int n, m;

struct Edge {
	int u, v;
	Edge() {}
	Edge(int u, int v) {
		this->u = u; this->v = v;
	}
} edge[M];

int en, first[N], next[M];

void add(int u, int v) {
	edge[en] = Edge(u, v);
	next[en] = first[u];
	first[u] = en++;
}

int h[505][505];
int pre[N], dfn[N], sccn, sccno[N], dfs_clock;
stack<int> S;

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = first[u]; i + 1; i = next[i]) {
		int v = edge[i].v;
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (dfn[u] == pre[u]) {
		sccn++;
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
	}
}

void find_scc() {
	sccn = dfs_clock = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 0; i < n * m; i++)
		if (!pre[i]) dfs_scc(i);
}

int in[N], out[N];

int main() {
	while (~scanf("%d%d", &m, &n)) {
		en = 0;
		memset(first, -1, sizeof(first));
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++)
				scanf("%d", &h[i][j]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; 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;
					if (h[i][j] >= h[x][y]) add(i * m + j, x * m + y);
				}
			}
		}
		find_scc();
		if (sccn == 1) {
			printf("0\n");
			continue;
		}
		memset(in, 0, sizeof(in));
		memset(out, 0, sizeof(out));
		for (int u = 0; u < n * m; u++) {
			for (int j = first[u]; j + 1; j = next[j]) {
				int v = edge[j].v;
				if (sccno[u] != sccno[v]) {
					in[sccno[v]]++;
					out[sccno[u]]++;
				}
			}
		}
		int ins = 0, outs = 0;
		for (int i = 1; i <= sccn; i++) {
			if (!in[i]) ins++;
			if (!out[i]) outs++;
		}
		printf("%d\n", max(ins, outs));
	}
	return 0;
}