[较难] UVa OJ 714 Copying books

题目描述

本题比较难，本人自己写了一份代码，但没有AC，下面给出《算法竞赛入门经典》书上代码，书上页码244

本题较难，仅Mark

// UVa714 Copying Books// Rujia Liu#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxm = 500 + 5;int m, k, p[maxm];// how many scribers needed if each scriber can work on at most maxp pagesint solve(long long maxp) {  long long done = 0;  int ans = 1;  for(int i = 0; i < m; i++) {    if(done + p[i] <= maxp) done += p[i];    else { ans++; done = p[i]; }  }  return ans;}int last[maxm]; // last[i] = 1 iff i is the last book assigned to someonevoid print(long long ans) {  long long done = 0;  memset(last, 0, sizeof(last));  int remain = k;  for(int i = m-1; i >= 0; i--) {    if(done + p[i] > ans || i+1 < remain) {      last[i] = 1; remain--; done = p[i];    }    else {      done += p[i];    }  }  for(int i = 0; i < m-1; i++) {    printf("%d ", p[i]);    if(last[i]) printf("/ ");  }  printf("%d\n", p[m-1]);}int main() {  int T;  scanf("%d", &T);  while(T--) {    scanf("%d%d", &m, &k);    long long tot = 0;    int maxp = -1;    for(int i = 0; i < m; i++) {      scanf("%d", &p[i]);      tot += p[i];      maxp = max(maxp, p[i]);    }    long long L = maxp, R = tot;    while(L < R) {      long long M = L + (R-L)/2;      if(solve(M) <= k) R = M; else L = M+1;    }    print(L);  }  return 0;}