二解 <书的复制(luoguP1281)>

题目详见:https://www.luogu.org/problem/show?pid=1281
本题有两种解法:
一.动态规划
类似于乘积最大那道题,关键是划分好书的分配方式,因为书本抄写是连续的,因此该问题是满足无后效性的,我们可以以抄书的人数为阶段,dp[i][j]表示前i个人抄写j本书需要抄写所消耗的最少时间.
动态规划转移方程为:dp[i][j] = min(dp[i][j],max(dp[i-1][k],s[j] - s[k]));其中1 <= k < j,s数组是前缀和, s[i]代表从1至i本书的页数和.
具体实现见代码:

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int n,m;int book[505],s[505];int dp[505][505]; //dp[i][j]代表前j本书由i个人复制的最优值 int ans[505][3];  //ans[i][1]用来存第i个人复制书稿的起点，ans[i][2]用来存第i个人复制书稿的终点。 void cal(int x){     int t = x, start,end = n;     int k=n;     for(int i = 1,j = m; i <= m; i++,j--)     {             ans[j][2] = end;              t = x;                         while(t-book[k] >=0 && k > 0)             {                    t -= book[k];                    k--;              }             ans[j][1] = k+1;             end = k;     }}int main(){    memset(dp,0x7f,sizeof(dp));    cin >> n >> m;    for(int i = 1; i<= n; i++)    {            cin >> book[i];            s[i] = s[i-1] + book[i];            dp[1][i] = s[i];    }    for(int i = 2; i<= m; i++)    {            for(int j = 1; j<= n; j++)            {                    int temp;                    for(int k = 1; k < j; k++)                    {                            temp = max(dp[i-1][k] , s[j] - s[k]);                            dp[i][j] = min(dp[i][j],temp);                    }               }    }    cal(dp[m][n]);    for(int i = 1; i <= m; i++)    {            cout << ans[i][1] <<" " << ans[i][2] << endl;    }    return 0;} 

二.二分答案
从题目可以分析,这道题的解一定是在1 ~ 所有书本的页数和s 之间,它们是线性且有序的,因些我们可以二分答案来解这道题.如果二分的mid能让抄书人从后向前分给m个人抄写能分得下,则让继续搜索左区间，继续寻找更小的值，如果m个人抄写还不够，则说明这个mid小了，需要搜索右区间。
具体代码见下面：

#include<iostream>using namespace std;int n,m;int book[505],s[505],ans[505][3],le,ri;int pd(int x){    int k = m,i = n,res = x;       while(k > 0)    {         res = x;        ans[k][2] = i;        while(i>0 && res - book[i] >= 0)        {            res -= book[i];            i --;               }       ans[k][1] = i+1;       k--;    }    if( i > 0)       return false;    else       return true;}void calAns(int x){    int k = m,i = n,res = x;       while(k > 0)    {         res = x;        ans[k][2] = i;        while(i>0 && res - book[i] >= 0)        {            res -= book[i];            i --;               }       ans[k][1] = i+1;       k--;    }}int main(){   cin >> n >> m;   for(int i = 1; i<= n; i++)   {       cin >> book[i];       s[i] = s[i-1] + book[i];   }   le = 0;ri = s[n] +1;   while(le+1 < ri)   {       int mid = le + (ri - le)/2;       if(pd(mid))       {            ri = mid;       }       else            le = mid;   }   calAns(ri) ;   for(int i = 1; i<= m; i++)       {           cout << ans[i][1] << " " << ans[i][2] << endl;       }   return 0;}