CF1005F Berland and the Shortest Paths

\(\color{#0066ff}{ 题目描述 }\)

一个无向图（边权为1），输出一下选边的方案使\(\sum d_i\)最小（\(d_i\)为从1到i的最短路）

输出一个方案数和方案（方案数超过k个只需输出k个）

\(\color{#0066ff}{输入格式}\)

第一行n，m，k，为点数，边数和k

接下来m行为无向边（权为1）

\(\color{#0066ff}{输出格式}\)

第一行为方案数（超过k个只输出k个）

接下来是01表示的方案（每条边选或不选）

\(\color{#0066ff}{输入样例}\)

4 4 3
1 2
2 3
1 4
4 3


    
4 6 3
1 2
2 3
1 4
4 3
2 4
1 3


    
    
5 6 2
1 2
1 3
2 4
2 5
3 4
3 5

\(\color{#0066ff}{输出样例}\)

2
1110
1011




1
101001


    
    
2
111100
110110

\(\color{#0066ff}{数据范围与提示}\)

\(n\leq 2*10^5, n-1\leq m \leq 2*10^5, 1 \leq k \leq 2*10^5, m*k\leq 10^6\)

\(\color{#0066ff}{ 题解 }\)

显然这题要求的是最短路图的生成树方案

先找出最短路图，根据dis（到1距离）分层

然后从2开始每个点都有向上连边的方案，统计一下

\(O(m*k)\)输出方案（搜索，枚举）

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
LL n, m, k;
const int maxn = 2e5 + 100;
struct node {
	int to, id;
	node *nxt;
	node(int to = 0, int id = 0, node *nxt = NULL): to(to), id(id), nxt(nxt) {}
	void *operator new(size_t) {
		static node *S = NULL, *T = NULL;
		return (S == T) && (T = (S = new node[1024]) + 1024), S++;
	}
};
using std::pair;
using std::make_pair;
std::priority_queue<pair<int, int>, std::vector<pair<int, int> >, std::greater<pair<int, int> > > q;
std::queue<int> v;
int cnt[maxn], dis[maxn], du[maxn];
bool vis[maxn];
node *head[maxn], *h[maxn];
void add(int from, int to, int id, node **hh) {
	hh[from] = new node(to, id, hh[from]);
}
void dij() {
	for(int i = 1; i <= n; i++) dis[i] = maxn;
	q.push(make_pair(dis[1] = 0, 1));
	while(!q.empty()) {
		int tp = q.top().second; q.pop();
		if(vis[tp]) continue;
		vis[tp] = true;
		for(node *i = head[tp]; i; i = i->nxt)
			if(dis[i->to] > dis[tp] + 1)
				q.push(make_pair(dis[i->to] = dis[tp] + 1, i->to));
	}
}
void toposort() {
	LL tot = 1;
	for(int i = 2; i <= n; i++) {
		tot *= du[i];
		if(tot >= k) break;
	}
	k = std::min(tot, k);
	printf("%lld\n", k);
}
void dfs(int d) {
	if(!k) return;
	if(d == n + 1) {
		for(int i = 1; i <= m; i++) printf("%d", vis[i]);
		puts("");
		k--;
		return;
	}
	for(node *i = h[d]; i; i = i->nxt) {
		vis[i->id] = true;
		dfs(d + 1);
		vis[i->id] = false;
		if(!k) return;
	}
}
int main() {
	n = in(), m = in(), k = in();
	int x, y;
	for(int i = 1; i <= m; i++) {
		x = in(), y = in();
		add(x, y, i, head);
		add(y, x, i, head);
	}
	dij();
	for(int i = 1; i <= n; i++)
		for(node *j = head[i]; j; j = j->nxt)
			if(dis[j->to] == dis[i] + 1)
				add(j->to, i, j->id, h), du[j->to]++;
	toposort();
	for(int i = 1; i <= m; i++) vis[i] = 0;
	dfs(2);
	return 0;
}