P2564 [SCOI2009]生日礼物 贪心

题意

有n个珠子,包括k中颜色,找出长度最小的一个区间,使得这个区间中包含所有的颜色。

思路

把n个珠子按照位子排序,然后维护每个珠子最右边能到的位子就行了。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

/*
        
⊂_ヽ
  \\ Λ_Λ  来了老弟
   \('ㅅ')
    > ⌒ヽ
   /   へ\
   /  / \\
   レ ノ   ヽ_つ
  / /
  / /|
 ( (ヽ
 | |、\
 | 丿 \ ⌒)
 | |  ) /
'ノ )  Lノ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/



            const int maxn = 1e6+9;
            pll a[maxn];
            ll d[100];
int main(){ 
            int n,k,tot=0;
            scanf("%d%d", &n, &k);
            for(int i=1; i<=k; i++){
                int num;    scanf("%d", &num);
                for(int j=1; j<=num; j++) {
                    int x;  scanf("%d", &x);
                    a[++tot].fi = x;
                    a[tot].se = i;
                }
            }
            sort(a +1, a+1+tot);
            ll ans = 1ll<<31;
            for(int i=1; i<=tot; i++){
                d[a[i].se] = a[i].fi;
                ll mx = 0, flag = 1, mn = 1ll<<31;

                for(int j=1; j<=k; j++){
                    mx = max(mx, d[j]);
                    mn = min(mn, d[j]);
                    if(d[j] == 0) flag = 0;
                }
                if(flag) ans = min(ans, mx - mn);
            }
            printf("%lld\n", ans);
            return 0;
}
View Code

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转载自www.cnblogs.com/ckxkexing/p/10404087.html
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