標簽： 最大獨立集 最小路徑覆蓋

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題目描述

In this problem, we would like to talk about unreachable sets of a directed acyclic graph G = (V, E). In mathematics a directed acyclic graph (DAG) is a directed graph with no directed cycles. That is a graph such that there is no way to start at any node and follow a consistently-directed sequence of edges in E that eventually loops back to the beginning again.
A node set denoted by V UR ? V containing several nodes is known as an unreachable node set of G if, for each two different nodes u and v in V UR , there is no way to start at u and follow a consistently-directed sequence of edges in E that ?nally archives the node v. You are asked in this problem to calculate the size of the maximum unreachable node set of a given graph G.

輸入

The input contains several test cases and the first line contains an integer T (1 ≤ T ≤ 500) which is the number of test cases.
For each case, the first line contains two integers n (1 ≤ n ≤ 100) and m (0 ≤ m ≤ n(n ? 1)/2) indicating the number of nodes and the number of edges in the graph G. Each of the following m lines describes a directed edge with two integers u and v (1 ≤ u, v ≤ n and u != v) indicating an edge from the u-th node to the v-th node. All edges provided in this case are distinct.
We guarantee that all directed graphs given in input are DAGs and the sum of m in input is smaller than 500000.

輸出

For each test case, output an integer in a line which is the size of the maximum unreachable node set of G.

樣例輸入

3 4 4 1 2 1 3 2 4 3 4 4 3 1 2 2 3 3 4 6 5 1 2 4 2 6 2 2 3 2 5

樣例輸出

2 1 3

題意

給定一個DAG，求出此圖滿足以下條件的最大點集合：集合中任意兩點相互不可達。

思路

先求出DAG的閉包，然后閉包的最大獨立集即為所求，DAG的閉包滿足性質
\[最大獨立集 = 最小路徑覆蓋 \]

• 然后我們可以把每個點拆分為兩個點，分別代表一個點的出度和入讀，那么會得到一個二分圖。
• 假設初始我們用\(n\)個路徑覆蓋所有的點，之后二分圖每增加一個匹配相當于把原DAG中的兩個路徑合并。所以\(最小路徑覆蓋數=n-二分圖最大匹配\)

代碼

// Ford-fulkerson算法 #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; typedef long long ll; const ll MOD=1e9+7; const int maxn=100050; const int inf=0x3f3f3f3f; int n,m; bool a[150][150]; struct Edge {int v,nxt,f,c; }e[maxn]; int h[maxn],tot=1; void addEdge(int x,int y,int w){e[++tot]=(Edge){y,h[x],0,w};h[x]=tot; } int s,t; bool vis[250]; int dfs(int x,int flow){if(vis[x]||flow==0) return 0;vis[x]=true;if(x==t){return flow;}int ans=0;for (int i = h[x]; i ; i=e[i].nxt){int t=min(flow,e[i].c-e[i].f);if(t>0&&(ans=dfs(e[i].v,t))){e[i].f+=ans;return ans;}t=min(flow,e[i^1].f);if(t>0&&(ans=dfs(e[i].v,t))){e[i^1].f-=ans;return ans;}}return 0; } int main(int argc, char const *argv[]) {int T;scanf("%d", &T);while(T--){scanf("%d%d", &n,&m);memset(a,false,sizeof a);memset(h,0,sizeof h);for (int i = 0; i < m; ++i){int x,y;scanf("%d%d", &x,&y);if(x>=1&&x<=n&&y>=1&&y<=n) a[x][y]=true;}tot=1;for (int k = 1; k <= n; ++k){for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){a[i][j]|=(a[i][k]&a[k][j]);}}}s=0,t=2*n+1;for (int i = 1; i <= n; ++i){addEdge(s,i,1);addEdge(i,s,0);addEdge(i+n,t,1);addEdge(t,i+n,0);for (int j = 1; j <= n; ++j){if(a[i][j]){addEdge(i,j+n,1);addEdge(j+n,i,0);}}}int mat=0,flow=0;memset(vis,false,sizeof vis);while(flow=dfs(s,inf)){mat+=flow;memset(vis,false,sizeof vis);}printf("%d\n", n-mat);}return 0; }

轉載于:https://www.cnblogs.com/sciorz/p/9021829.html

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