2017 ICPC 南宁 M题 The Maximum Unreachable Node Set (二分图匹配)
The Maximum Unreachable Node Set
时间限制: 20 Sec 内存限制: 128 MB提交: 51 解决: 8
[提交][状态][讨论版][命题人:admin]
题目描述
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 finally 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.
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 finally 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.
输入
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 6= 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.
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 6= 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.
输出
For each test case, output an integer in a line which is the size of the maximum unreachable node set of G.
样例输入
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
样例输出
2
1
3
给你n个点m条边的DAG,问最多选多少边,使得集合内地点都不可达。
可以用Floyd得到可达矩阵,然后二分匹配跑一下最大匹配,不就是求出了两两相连的最多的点。然后n减去这些点,就是结果。水的要命。
代码实现:
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<vector> #include<map> #include<set> #include<cstdio> #define ll long long #define mset(a,x) memset(a,x,sizeof(a)) using namespace std; const double PI=acos(-1); const int inf=0x3f3f3f3f; const double esp=1e-6; const int maxn=1e6+5; const int mod=1e9+7; int dir[4][2]={0,1,1,0,0,-1,-1,0}; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;} ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;} ll Fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n;n=n*n;}return r;} ll upd(ll x,ll v){x=x+v>=mod?x+v-mod:x+v;return x;} int s[105][105],n,m,vis[105],link[105]; int dfs(int num) { for(int i=1;i<=n;i++) { if(s[num][i]&&!vis[i]) { vis[i]=1; if(link[i]==-1||dfs(link[i])) { link[i]=num; return 1; } } } return 0; } int main() { int i,j,k,t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); mset(s,0); while(m--) { int x,y; scanf("%d%d",&x,&y); s[x][y]=1; } //求可达矩阵 for(k=1;k<=n;k++) { for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(s[i][k]&&s[k][j]) s[i][j]=1; } } } mset(link,-1); int ans=0; for(i=1;i<=n;i++) { mset(vis,0); if(dfs(i)) ans++; } printf("%d\n",n-ans); } }