POJ 3312 Mahershalalhashbaz, Nebuchadnezzar, and Billy Bob Benjamin Go to the Regionals (水题,贪心)
Go to The and poj 贪心 水题
2023-09-11 14:17:18 时间
题意:给定 n 个字符串,一个k,让你把它们分成每组k个,要保证每组中每个字符串长度与它们之和相差不能超2.
析:贪心策略就是长度相差最小的放上块。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <functional> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <deque> #include <map> #include <cctype> #include <stack> #include <sstream> using namespace std ; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 5; const int mod = 1e9 + 7; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; int n, m; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } char s[maxn]; vector<int> v; int main(){ int kase = 0; while(scanf("%d %d", &n, &m) == 2){ if(!m && !n) break; v.clear(); for(int i = 0; i < n; ++i){ scanf("%s", s); v.push_back(strlen(s)); } sort(v.begin(), v.end()); bool ok = true; for(int i = 0; i < n/m; i++){ double cnt = 0.0; for(int j = 0; j < m; ++j) cnt += v[i*m+j]; cnt /= (double)m*1.0; for(int j = 0; j < m; ++j) if(abs((double)v[i*m+j]*1.0 - cnt) > 2.0){ ok = false; break; } } if(kase++) printf("\n"); printf("Case %d: %s\n", kase, ok ? "yes" : "no"); } return 0; }
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