标签:com 1.0 fun iostream bitset vector char msu space
题目大意:
nnn个位置,每个位置有两个属性s,cs,cs,c,要求选择3个位置i,j,ki,j,ki,j,k,使得si<sj<sks_i<s_j<s_ksi?<sj?<sk?,并使得ci+cj+ckc_i+c_j+c_kci?+cj?+ck?最小
输入格式:
一行一个整数,nnn,3<=n<=30003<=n<=30003<=n<=3000
一行nnn个整数,即sss
再一行nnn个整数,即ccc
输出格式:
输出一个整数,即最小的c_i+c_j+c_k
枚举 j ,分段暴力;
O(N^2);
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 900005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-4 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == ‘-‘) f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int n; ll s[3006]; ll c[3006]; ll minn1[4000]; ll minn2[4000]; int main() { //ios::sync_with_stdio(0); rdint(n); for (int i = 1; i <= n; i++)rdllt(s[i]); for (int i = 1; i <= n; i++)rdllt(c[i]); ll ans = inf; c[0] = c[n + 1] = inf; for (int i = 2; i < n; i++) { int l = 0, r = n + 1; for (int j = 1; j < i; j++) { if (s[j] < s[i]) if (c[j] < c[l])l = j; } for (int j = i + 1; j <= n; j++) { if (s[j] > s[i]) if (c[j] < c[r])r = j; } if (l != 0 && r != n + 1)ans = min(ans, c[i] + c[l] + c[r]); } if (ans == inf)cout << -1 << endl; else cout << ans << endl; return 0; }
标签:com 1.0 fun iostream bitset vector char msu space
原文地址:https://www.cnblogs.com/zxyqzy/p/10273592.html