「TJOI2018」智力竞赛

题解
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

#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) 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; 
} 
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) l = mid + 1,ans = mid; 
		else r = mid - 1; 
	} 
	if(l <= mx) print(ans),pc('\n'); 
	else puts("AK"); 
	return 0; 
}

标签：10,include,int,pc,maxn,gc,TJOI2018,智力竞赛
来源： https://www.cnblogs.com/cutemush/p/12815205.html