「TJOI2018」智力竞赛
题解好难看啊。。。
就是求可重路径覆盖之后最大化剩余点的最小权值
二分答案后就是一个可重复路径覆盖
处理出可达点做二分图匹配就好了
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 | #include<cstdio> #include<cstring> #include<algorithm> #define gc getchar() #define pc putchar inline int read() { int x = 0,f = 1; char c = getchar (); while (c < '0' || c > '9' ) c = gc; while (c <= '9' && c >= '0' ) x = x * 10 + c - '0' ,c = gc; return x * f; } void print( int x) { if (x < 0) { pc( '-' ); x = -x; } if (x >= 10) print(x / 10); pc(x % 10 + '0' ); } const int maxn = 507; int n,m; bool mp[maxn][maxn]; int val[maxn]; int a[maxn]; void floyd() { for ( int k = 1;k <= n;++ k) for ( int i = 1;i <= n;++ i) for ( int j = 1;j <= n;++ j) mp[i][j] |= mp[i][k] & mp[k][j]; } int vis[maxn]; int tot = 0; int bel[maxn]; bool find( int x, int fa) { for ( int i = 1;i <= tot;++ i) { if (vis[i] != fa && mp[a[x]][a[i]]) { vis[i] = fa; if (!bel[i] || find(bel[i],fa)) { bel[i] = x; return true ; } } } return false ; } int check( int x) { tot = 0; for ( int i = 1;i <= m;++ i) if (val[i] < x) //统计有效点有多少个 a[++ tot] = i; int ret = tot; memset (bel,0, sizeof bel); for ( int i = 1;i <= tot;++ i) { if (find(i,i)) ret --; } return ret; //返回最小路径覆盖的值 } int main() { //freopen("contest2.in","r",stdin); n = read() + 1, m = read(); int mx = 0; for ( int k,i = 1;i <= m;++ i) { val[i] = read(); mx = std::max(mx,val[i]); k = read(); for ( int v,j = 1;j <= k;++ j) { v = read(); mp[i][v] = 1; } } floyd(); int ans = -1; int l = 1,r = mx; while (l <= r) { int mid = l + r >> 1; if (check(mid) <= n) //如果覆盖成功的话,则左边界可以再增加 l = mid + 1,ans = mid; else r = mid - 1; } if (l <= mx) print(ans),pc( '\n' ); else puts ( "AK" ); return 0; } |
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步