POJ 2375 Cow Ski Area 强连通图最少添边数

来源：互联网发布：elliott smith 知乎编辑：程序博客网时间：2024/05/12 09:29

题目来源：POJ 2375 Cow Ski Area

题意：n*m的一个矩阵如果在一个点上可以往上下左右4个点并且必须大于等于要去的点可以安装一个任意门每个任意门可以连接容易2个点

求最少安装多少个任意门可以是全通连通

思路：强连通分量缩点求出度为0和入读为0的连通分量2者的最大值特判全图强连通的情况

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int maxn = 510;int first[maxn*maxn];int next[maxn*maxn];stack <int> s;int pre[maxn*maxn];int low[maxn*maxn];int sccno[maxn*maxn];//每个点所在强联通分量的scc_cnt int in[maxn*maxn];//每个强联通分量的入度 int out[maxn*maxn];int dfs_clock;int scc_cnt;int n, m, nm;int a[maxn][maxn];int cnt;int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};struct edge{    int v, next;}e[maxn*maxn*4];void ad(int x, int y){    e[cnt].next = first[x];    e[cnt].v = y;    first[x] = cnt++;}void dfs(int u){pre[u] = low[u] = ++dfs_clock;s.push(u);for(int ee = first[u]; ee != -1; ee = e[ee].next){int v = e[ee].v;if(!pre[v]){dfs(v);low[u] = min(low[u], low[v]);}else if(!sccno[v]){low[u] = min(low[u], pre[v]);}}if(low[u] == pre[u]){scc_cnt++;while(1){int x = s.top();s.pop();sccno[x] = scc_cnt;if(x == u)break;}}}void find_scc(){dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for(int i = 0; i < nm; i++)if(!pre[i])dfs(i);}void ok(int x, int y){for(int i = 0; i < 4; i++){int xx = x + dir[i][0];int yy = y + dir[i][1];if(xx < 0 || xx >= n || yy < 0 || yy >= m)continue;if(a[x][y] >= a[xx][yy]){ad(x*m+y, xx*m+yy);}}}int main(){scanf("%d %d", &m, &n);nm = n*m;memset(first, -1, sizeof(first));memset(next, -1, sizeof(next));for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)scanf("%d", &a[i][j]);cnt = 1;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){ok(i, j);}}find_scc();memset(in, 0, sizeof(in));memset(out, 0, sizeof(out));for(int u = 0; u < nm; u++){for(int ee = first[u]; ee != -1; ee = e[ee].next){int v = e[ee].v;if(sccno[u] != sccno[v]){in[sccno[v]]++;out[sccno[u]]++;}}}int sum1 = 0, sum2 = 0;for(int i = 1; i <= scc_cnt; i++){if(!in[i])sum1++;if(!out[i])sum2++;}int ans = max(sum1, sum2);if(scc_cnt == 1)ans = 0;printf("%d\n", ans);return 0;}

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