bzoj3624: [Apio2008]免费道路

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[TOC]

#题目链接
bzoj3624: [Apio2008]免费道路
#题解
两边kruskal分别找出必须保留的石子路和软石路
剩下的随便填满k就ok了
#代码

#include<map> 
#include<vector> 
#include<cstdio> 
#include<algorithm> 
#define gc getchar()
#define pc putchar 
inline int read() { 
	int x = 0,f = 1; 
	char c = gc; 
	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 = 2000007; 
int a[maxn]; 
int n,m,k; 
struct node { 
	int u,v,w,tag; 
} e[maxn]; 
int num = 0; 
int fa[maxn]; 
inline void add(int a,int b,int c) { 
	e[++ num].u = a,e[num].v = b; e[num].w = c; 
} 
inline bool c1(node a,node b) {
	return a.w < b.w; 
} 
inline bool c2(node a,node b) { 
	return a.w > b.w; 
} 
int find(int x) { 
	if(x != fa[x]) fa[x] = find(fa[x]); 
	return fa[x]; 
} 
int main() {
	n = read(),m = read(),k = read(); 
	for(int u,v,w,i = 1;i <= m;++ i) { 
		u = read(),v = read(),w = read(); 
		add(u,v,w); 
	} 
	for(int i = 1;i <= n;++ i) fa[i] = i; 
	std::sort(e + 1,e + m + 1,c2); 
	int cnt = 0; 
	for(int i = 1;i <= m;++ i)  {
		int fx = find(e[i].u),fy = find(e[i].v); 
		if(fx != fy) {
			fa[fx] = fy; 
			if(!e[i].w) e[i].tag = 1,cnt ++; 
		} 
	}  
	if(cnt > k) { 
		puts("no solution"); 
		return 0; 
	} 	 
	for(int i = 1;i <= n;++ i) fa[i] = i; 
	for(int i = 1;i <= m;++ i) 
		if(e[i].tag) fa[find(e[i].u)] = fa[find(e[i].v)]; 
	std::sort(e + 1,e + m + 1,c1) ; 
	for(int i = 1;i <= m;++ i) { 
		if(e[i].w) break; 
		int fx = find(e[i].u),fy =  find(e[i].v); 
		if(fx != fy) { 
			fa[fx] = fy; 
			++ cnt; 
			e[i].tag = 1; 
		} 
		if(cnt == k) break; 
	} 
	if(cnt < k) { 
		puts("no solution"); 
		return 0; 
	} 
	std::sort(e + 1,e + m + 1,c2); 
	for(int i = 1;i <= m;++ i)  { 
		if(!e[i].w) break; 
		int fx = find(e[i].u),fy = find(e[i].v); 
		if(fx != fy) { 
			fa[fx] = fy; 
			e[i].tag = 1; 
		} 
	} 
	for(int i = 1;i <= m;++ i) 
		if(e[i].tag) { 
			print(e[i].u);pc(' '); 
			print(e[i].v);pc(' '); 
			print(e[i].w);pc('\n'); 
			
		} 
	return 0; 
}