[SCOI2009] birthday gift monotonic foot emulated

The meaning of problems: give you the n k colors, each point has two coordinates and color attributes, select a length of the interval as short as possible, so that the point of each color are present in the interval.

 

data range:

For 50% of the data, N≤10000;

For 80% of the data, N≤800000;

To 100% of the data, 1≤N≤1000000,1≤K≤60, 0 ≦ beads position <2 ^ {31} .

-------------------------------------------------- I'm dividing line ---------------------------------------------- ---------

Solution: monotonic foot emulated classic application. The coordinate point of first order, two variableslandrto enumerate interval,ifltorinterval does not satisfy the requirements,r++,ifltorsections meet, the answer records,l ++.

Sort time complexity O (NlogN), foot takes time complexity O (N), the total time complexity is O (NlogN).

#include<bits/stdc++.h>

#define ll long long
#define mp make_pair
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)

using namespace std;

typedef pair<int, int> pii;
typedef double db;
const int N = 1e6 + 50;
struct node { int x, id; } a[N]; 
int n, k, h, vis[N], q[N], u, ans = 1e9, cnt; 
inline int read(){
    int x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar();}
    while(ch >='0' && ch <='9') { x = (x<<3)+(x<<1)+(ch^48); ch = getchar();}
    return x*f;
}
bool mycmp(node a, node b){ return a.x < b.x; }
void insert(int x) { if(vis[x] == 0) cnt++; vis[x]++; }
void remove(int x) { if(vis[x] == 1) cnt--; vis[x]--; }
void init(){
    n = read(); k = read();
    rep(i, 1, k){
        int t = read();
        rep(j, 1, t) a[++h].x = read(), a[h].id = i;
    }
    sort(a+1, a+n+1, mycmp);
}
void work(){
    int l = 1, r = 1;
    while(r <= n){
        insert(a[r].id);
        while(true) {
            remove(a[l].id); 
            if(cnt == k) l++;
            else { insert(a[l].id); break;}
        }
        if(cnt == k) ans = min(ans, a[r].x - a[l].x);
        r++;
    }
    printf("%d\n", ans);
}
int main(){
    init();
    work();
    return 0;
}

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