匈牙利算法和hopcroft_karp算法模板

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今晚学习了hopcroft_karp算法，留个模板，顺便也留个匈牙利算法模板。

题目为POJ 1469，在本题中，两种算法所用时间基本一致

//hopcroft_karp算法，复杂度O(sqrt(n)*m)#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;const int N = 320;const int INF = 0x3f3f3f3f;struct edge{    int to, next;}g[N*N];int match[N], head[N];bool used[N];int p, n;int nx, ny, cnt, dis; //nx,ny分别是左点集和右点集的点数int dx[N], dy[N], cx[N], cy[N]; //dx,dy分别维护左点集和右点集的标号//cx表示左点集中的点匹配的右点集中的点，cy正好相反void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}bool bfs() //寻找增广路径集，每次只寻找当前最短的增广路{    queue<int> que;    dis = INF;    memset(dx, -1, sizeof dx);    memset(dy, -1, sizeof dy);    for(int i = 1; i <= nx; i++)        if(cx[i] == -1) //将未遍历的节点入队，并初始化次节点距离为0            que.push(i), dx[i] = 0;    while(! que.empty())    {        int v = que.front(); que.pop();        if(dx[v] > dis) break;        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(dy[u] == -1)            {                dy[u] = dx[v] + 1;                if(cy[u] == -1) dis = dy[u]; //找到了一条增广路，dis为增广路终点的标号                else                    dx[cy[u]] = dy[u] + 1, que.push(cy[u]);            }        }    }    return dis != INF;}int dfs(int v){    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(! used[u] && dy[u] == dx[v] + 1) //如果该点没有被遍历过并且距离为上一节点+1        {            used[u] = true;            if(cy[u] != -1 && dy[u] == dis) continue; //u已被匹配且已到所有存在的增广路终点的标号，再递归寻找也必无增广路，直接跳过            if(cy[u] == -1 || dfs(cy[u]))            {                cy[u] = v, cx[v] = u;                return 1;            }        }    }    return 0;}int hopcroft_karp(){    int res = 0;    memset(cx, -1, sizeof cx);    memset(cy, -1, sizeof cy);    while(bfs())    {        memset(used, 0, sizeof used);        for(int i = 1; i <= nx; i++)            if(cx[i] == -1)                res += dfs(i);    }    return res;}int main(){    int t, a, b;    scanf("%d", &t);    while(t--)    {        cnt = 0;        memset(head, -1, sizeof head);        scanf("%d%d", &p, &n);        for(int i = 1; i <= p; i++)        {            scanf("%d", &a);            for(int j = 0; j < a; j++)            {                scanf("%d", &b);                add_edge(i, b);            }        }        nx = p, ny = n;        if(hopcroft_karp() == p) printf("YES\n");        else printf("NO\n");    }    return 0;}

//匈牙利算法，复杂度O(nm)#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>using namespace std;const int N = 320;struct edge{    int to, next;}g[N*N];int match[N], head[N];bool use[N];int p, n;int nx, ny, cnt;void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}bool dfs(int v){    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(use[u] == false)        {            use[u] = true;            if(match[u] == -1 || dfs(match[u]))            {                match[u] = v;                return true;            }        }    }    return false;}int hungary(){    int res = 0;    memset(match, -1, sizeof match);    for(int i = 1; i <= nx; i++)    {        memset(use, 0, sizeof use);        if(dfs(i)) res++;    }    return res;}int main(){    int t, a, b;    scanf("%d", &t);    while(t--)    {        cnt = 0;        memset(head, -1, sizeof head);        scanf("%d%d", &p, &n);        for(int i = 1; i <= p; i++)        {            scanf("%d", &a);            for(int j = 0; j < a; j++)            {                scanf("%d", &b);                add_edge(i, b);            }        }        nx = p, ny = n;        if(hungary() == p) printf("YES\n");        else printf("NO\n");    }    return 0;}

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