UVA - 714 Copying Books (抄书)(二分+贪心)

题意:把一个包含m个正整数的序列划分成k个(1<=k<=m<=500)非空的连续子序列,使得每个正整数恰好属于一个序列(所有的序列不重叠,且每个正整数都要有所属序列)。设第i个序列的各数之和为S(i),你的任务是让所有的S(i)的最大值尽量小。如果有多解,S(1)应尽量小,如果仍有多解,S(2)应尽量小,依此类推。

分析:

1、二分最小值x。

2、判断当前x是否满足条件时,从右往左尽量划分,若cnt<k,则从0开始依次标为分界点,这样可满足S(1),S(2),……,尽量小。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 500 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL a[MAXN];
int m, k;
int cnt;
int vis[MAXN];//分界点
bool judge(LL x){
    memset(vis, 0, sizeof vis);
    cnt = 0;
    int pos = m - 1;
    while(pos >= 0){
        int num = 0;//当前序列中的元素
        LL sum = 0;
        while(pos >= 0 && sum + a[pos] <= x){
            sum += a[pos];
            ++num;
            --pos;
        }
        if(!num) return false;//若当前序列中没有元素,说明x太小
        if(pos >= 0) vis[pos] = 1;
        ++cnt;
    }
    if(cnt > k) return false;
    return true;
}
void solve(){
    LL l = 0, r = 1e15;
    while(l < r){
        LL mid = l + (r - l) / 2;
        if(judge(mid)) r = mid;
        else l = mid + 1;
    }
    if(judge(r)){
        for(int i = 0; i < m - 1 && cnt < k; ++i){
            if(!vis[i]){
                vis[i] = 1;
                ++cnt;
            }
        }
        for(int i = 0; i < m; ++i){
            if(i) printf(" ");
            printf("%lld", a[i]);
            if(vis[i]) printf(" /");
        }
        printf("\n");
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &m, &k);
        for(int i = 0; i < m; ++i){
            scanf("%lld", &a[i]);
        }
        solve();
    }
    return 0;
}

  

posted @ 2017-02-06 18:06  Somnuspoppy  阅读(320)  评论(0编辑  收藏  举报