POJ 3312 Mahershalalhashbaz, Nebuchadnezzar, and Billy Bob Benjamin Go to the Regionals (水题,贪心)

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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