jzoj5177. 【NOIP2017提高组模拟6.28】TRAVEL(二分)

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5177. 【NOIP2017提高组模拟6.28】TRAVEL

Description

Input

Output

Sample Input
4 4
1 2 1 10
2 4 3 5
1 3 1 5
2 4 2 7

Sample Output
6
2 3 4 5 6 7

分析：枚举区间起点，二分终点，然后每条边都走一次判断是否能走到n。

代码

#pragma optimize(2)
#include <cstdio>
#include <algorithm>
#define N 10000
using namespace std;

struct arr
{
	int l,r,nxt,to,num;
}a[N];
struct node
{
	int l,r,x,y;
}b[N];
int n,m,tot,p,ls[N],v[N];
bool fl;

int cmp(node i, node j){return i.r < j.r;}

inline void add(int x, int y, int l, int r, int i)
{
	a[++tot].to = y;
	a[tot].nxt = ls[x];
	a[tot].l = l;
	a[tot].r = r;
	a[tot].num = i;
	ls[x] = tot;
}

inline int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    } while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    } return x * f;
}


inline void dfs(int x, int L, int R)
{
	if (fl) return;
	if (x == n){fl = true; return;}
	for (int i = ls[x]; i; i = a[i].nxt)
		if (v[a[i].num] != p && a[i].l <= L && a[i].r >= R)
		{
			v[a[i].num] = p;
			dfs(a[i].to, L, R);
		}
}

int main()
{
	int I, J;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++)
	{
		int l,r,x,y;
		x=read(),y=read(),l=read(),r=read();
		add(x, y, l, r, i);
		add(y, x, l, r, i);	
		b[i].x = x;
		b[i].y = y;
		b[i].l = l;
		b[i].r = r;
	}
	sort(b + 1, b + m + 1, cmp);
	int ans = 0;
	for (int i = 1; i <= m; i++)
	{
		int L = 1, R = m, j = 0;
		while (L <= R)
		{
			int mid = (L + R) / 2;
			fl = false;
			p++; 
			dfs(1, b[i].l, b[mid].r);
			if (fl) j = mid, L = mid + 1;
				else R = mid - 1;
		}
		if (b[j].r - b[i].l + 1 > ans)
		{
			ans = b[j].r - b[i].l + 1;
			J = b[j].r;
			I = b[i].l;
		}
		if (b[j].r - b[i].l + 1 == ans) if (b[i].l < I) I = b[i].l, J = b[j].r;
	}
	printf("%d\n", ans);
	if (ans) 
		for (int i = I; i <= J; i++) printf("%d ", i);
}