POJ

最小边覆盖问题 也是匹配问题

这里用到了 匈牙利算法，

还用到了结构体数组存邻接表的做法

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>#include <stack>#include <queue>#include <ctype.h>#include <vector>#include <algorithm>#include <sstream>#define PI acos(-1.0)#define in freopen("in.txt", "r", stdin)#define out freopen("out.txt", "w", stdout)using namespace std;typedef long long ll;const int maxn = 1500+7, INF = 0x3f3f3f3f;int n, cur, ans;int f[maxn];bool vis[maxn];int next[maxn];struct edge {    int v, bef;}e[maxn*2];void add(int x, int y) {    e[cur].v = y;    e[cur].bef = next[x];    next[x] = cur++;}void init() {    memset(next, -1, sizeof(int)*(n+1));    cur = 1;    for(int j = 0; j < n; ++j) {        int u, v, cnt;        scanf("%d:(%d)", &u, &cnt);        for(int i = 0; i < cnt; ++i) {            scanf("%d", &v);            add(u, v);            add(v, u);            //a[u].push_back(v);            //a[v].push_back(u);        }    }    memset(f, -1, sizeof(int)*(n+1));}bool dfs(int id) {    for(int i = next[id]; i != -1; i = e[i].bef) {        if(!vis[e[i].v]) {            vis[e[i].v] = 1;            if(f[e[i].v] == -1 || dfs(f[e[i].v])) {                f[id] = e[i].v;                f[e[i].v] = id;                return true;            }        }    }    return false;}int main() {    while(scanf("%d", &n) != EOF && n) {        init();        ans = 0;        for(int i = 0; i < n; ++i) {            if(f[i] == -1) {                memset(vis, false, sizeof(bool)*(n+1));                if(dfs(i)) ans++;            }        }        printf("%d\n", ans);    }    return 0;}