c/c++语言开发共享洛谷P4589 [TJOI2018]智力竞赛(二分答案 二分图匹配)

题意

题目链接

给出一个带权有向图，选出n + 1n+1条链，问能否全部点覆盖，如果不能，问不能覆盖的点权最小值最大是多少

sol

tjoi怎么净出板子题

二分答案之后直接二分图匹配check一下。

多读读题就会发现题目要求的就是可相交的最小路径覆盖，那么按照套路先floyd一遍，如果能联通的话就再二分图中加边，然后判一下最大匹配数就行了。刚开始以为因为有的点可以不选，要在匈牙利的时候进行玄学贪心，其实是不用的，因为我们已经求过传递闭包了。所以直接求就是对的

因为(m leqslant 500)，所以floyd的时候要用bitset优化一下

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second //#define int long long  #define ull signed long long  #define ll long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1001, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, m, a[maxn], ans[maxn], vis[maxn], tim = 1, link[maxn], st[maxn], top; bitset<maxn> can[maxn]; bool dfs(int x) {     for(int i = 1; i <= top; i++) {         if(can[st[x]][st[i]]) {             if(vis[i] == tim) continue;             vis[i] = tim;             if(!link[i] || (dfs(link[i]))) {link[i] = x; return 1;}         }     }     return 0; } bool check(int val) {     memset(link, 0, sizeof(link)); top = 0;     for(int i = 1; i <= m; i++) if(a[i] < val) st[++top] = i;     int ans = 0;     for(int i = 1; i <= top; i++, tim++) if(dfs(i)) ans++;     return top - ans <= n + 1; } signed main() {     n = read(); m = read();      int mx = 0;     for(int i = 1; i <= m; i++) {         a[i] = read(); int k = read(); mx = max(a[i], mx);         for(int j = 1; j <= k; j++) can[i][read()] = 1;     }     for(int k = 1; k <= m; k++)         for(int i = 1; i <= m; i++)             if(can[i][k])                 can[i] |= can[k];     int l = 0, r = mx, ans = 0;     while(l <= r) {         int mid = l + r >> 1;         if(check(mid)) ans = mid, l = mid + 1;         else r = mid - 1;     }     if(ans >= mx) puts("ak");     else cout << ans;     return 0; } /*  */