题目链接：
传送门
题目大意：
Bob非常享受玩电脑游戏的过程，尤其是策略游戏，但是在有些时候，他因为不能在第一时间找到最佳的策略而十分伤心。 现在，他遇到了一个问题。他必须保卫一个中世纪的城市，有很多道路将整个城市连起来，整体上看上去像一棵树。Bob需要放置尽可能少的士兵，保卫树上所有的边。士兵只能放在节点上，但是却可以保卫所有与这个节点相邻的边。
题目分析：
由于树本身具有二分图的性质，所以这里我把树手动染色转成二分图，然后跑最小点覆盖
（当然也可以树形dp）
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N (1500 + 5)
using namespace std;
inline int read(){int cnt = 0, f = 1; char c;c = getchar();while (!isdigit(c)) {if(c == '-') f = -f;c = getchar();}while (isdigit(c)) {cnt = cnt * 10 + c - '0';c = getchar();}return cnt * f;
}
int n, nxt[2*N], to[2*N], first[2*N], tot, x, t, y, Float[N], cnt, black, white, ans, match[N];
bool vis[N];
struct node{int x;int y;
}edge[N];
void add(int x, int y) {nxt[++tot] = first[x];first[x] = tot;to[tot] = y;
}
void Dfs(int u){for (register int i = first[u]; i; i = nxt[i]) {int v = to[i];if(!Float[v]) {Float[v] = Float[u] + 1;Dfs(v);}}
}
void build_map() {memset (first, 0, sizeof(first));memset (nxt, 0, sizeof(nxt));memset (to, 0, sizeof(to));tot = 0;for (register int i = 1; i <= cnt; i++) {int X = edge[i].x; int Y = edge[i].y;if (Float[X] % 2 == 1) {add(X, Y);if (!vis[X]) {vis[X] = true;black++;}if (!vis[Y]) {vis[Y] = true;white++;}} else {add(Y, X);if (!vis[X]) {vis[X] = true;white++;}if (!vis[Y]) {vis[Y] = true;black++;}}}
}
int find(int u) {
// cout<<u<<"##"<<endl;for (register int i = first[u]; i; i = nxt[i]) {int v = to[i];
// cout<<vis[v]<<" "<<v<<endl;if (vis[v]) {continue; } else {vis[v] = true;if (match[v] == -1||find(match[v])) {match[v] = u;return 1;}}}return 0;
}int hungary() {for (register int i = 1; i <= n; i++) match[i] = -1;for (register int i = 1; i <= n; i++) {for (register int j = 1; j <= n; j++) vis[j] = 0;ans += find(i);}return ans;
}int main(){while(scanf("%d", &n) != EOF) {memset(first, 0, sizeof(first));memset(nxt, 0, sizeof(nxt));memset(to, 0, sizeof(to));memset(Float, 0, sizeof(Float));memset(vis, 0, sizeof(vis));tot=0; cnt=0; ans=0;for (register int i = 1; i <= n; i++) {x = read();t = read();for (register int j = 1; j <= t; j++) {y = read();edge[++cnt].x = x+1;edge[cnt].y = y+1;add(x+1,y+1);add(y+1,x+1);}}Float[1] = 1;Dfs(1);build_map();
// cout<<black<<" "<<white<<endl;
// for(register int i = 1; i <= n; i++) cout<<Float[i];int res = hungary();printf("%d\n", res);}return 0;
}
