• 最小路径覆盖问题 最大流


    最小路径覆盖数=顶点数-最大匹配数;

    找路径的时候,回溯找;

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 200005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-5
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    
    inline int rd() {
    	int x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    
    int n, m;
    int st, ed;
    int fa[maxn];
    void init() {
    	for (int i = 0; i <= 2 * n + 2; i++)fa[i] = i;
    }
    struct node {
    	int u, v, nxt, w;
    }edge[maxn << 1];
    
    int head[maxn], cnt;
    
    void addedge(int u, int v, int w) {
    	edge[cnt].u = u; edge[cnt].v = v; edge[cnt].nxt = head[u];
    	edge[cnt].w = w; head[u] = cnt++;
    }
    
    int rk[maxn];
    
    int bfs() {
    	queue<int>q;
    	ms(rk);
    	rk[st] = 1;
    	q.push(st);
    	while (!q.empty()) {
    		int tmp = q.front(); q.pop();
    		for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
    			int to = edge[i].v;
    			if (rk[to] || edge[i].w <= 0)continue;
    			rk[to] = rk[tmp] + 1; q.push(to);
    		}
    	}
    	return rk[ed];
    }
    
    int dfs(int u, int flow) {
    	if (u == ed)return flow;
    	int add = 0;
    	for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
    		int v = edge[i].v;
    		if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
    		int tmpadd = dfs(v, min(edge[i].w, flow - add));
    		if (!tmpadd) { rk[v] = -1; continue; }
    		edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd;
    		add += tmpadd;
    	}
    	return add;
    }
    
    int ans;
    void dinic() {
    	while (bfs())ans += dfs(st, inf);
    }
    int findfa(int x) {
    	if (x == fa[x]) {
    		return x;
    	}
    	return fa[x] = findfa(fa[x]);
    }
    int vis[maxn];
    void output(int x) {
    	printf("%d ", x);
    	for (int i = head[x]; i != -1; i = edge[i].nxt) {
    		if (edge[i].w == 0 && edge[i].v > n)output(edge[i].v - n);
    	}
    }
    int main()
    {
    	//ios::sync_with_stdio(0);
    	memset(head, -1, sizeof(head));
    	
    	n = rd(); m = rd(); st = 0; ed = 2 * n + 1;init();
    	for (int i = 1; i <= m; i++) {
    		int u, v;
    		u = rd(); v = rd();
    		addedge(u, v + n, 1); addedge(v + n, u, 0);
    	}
    	for (int i = 1; i <= n; i++) {
    		addedge(st, i, 1); addedge(i, st, 0);
    	}
    	for (int i = 1; i <= n; i++) {
    		addedge(i + n, ed, 1); addedge(ed, i + n, 0);
    	}
    	dinic();
    	for (int i = 0; i < cnt; i++) {
    		if (edge[i].u >= 1 && edge[i].u <= n && edge[i].v > n&& edge[i].w == 0) {
    			fa[findfa(edge[i].v - n)] = findfa(edge[i].u);
    		}
    	}
    	for (int i = 1; i <= n; i++) {
    		if (findfa(i) == i) {
    			output(i); printf("
    ");
    		}
    		
    	}
    	
    	cout << n-ans << endl;
    	return 0;
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10349789.html
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