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

时间：2021-10-05 02:08:57

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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