POJ 1416.Shredding Company

来源：互联网发布：艾滋病歧视知乎日报编辑：程序博客网时间：2024/05/29 19:24

题目:http://poj.org/problem?id=1416

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <map>#include <math.h>#include <string>#include <string.h>#include <bitset>#define INF 0xfffffff#define MAXN 100105using namespace std;int t,cnt;string num;string part[10];string ans[10];int len;int anscnt, anssum;bool rj;int isOK(){int sum = 0;for(int i = 0; i < cnt; i++){int tmp = atoi(part[i].c_str());sum+=tmp;}if(sum<=t)return sum;else return -1;}void dfs(int pos){if(pos>=len){int sum = isOK();if(sum!=-1){if(sum>anssum){for(int i = 0; i < cnt; i++)ans[i] = part[i];anscnt = cnt;anssum = sum;rj = false;}else if(sum==anssum){rj = true;}}return;}int subpos = pos;for(int i = 0; i < cnt; i++){if(part[i].length()<subpos)subpos -= part[i].length();else{dfs(pos+1);string cur[10];for(int k = 0; k < cnt; k++)cur[k] = part[k];cnt++;for(int k = cnt-1; k >= i+2; k--)part[k] = part[k-1];part[i+1] = part[i].substr(subpos,part[i].length());part[i] = part[i].substr(0,subpos);dfs(pos+1);cnt--;for(int k = 0; k < cnt; k++)part[k] = cur[k];}}}int main(){while(cin>>t>>num){if(t==0&&num=="0")break;part[0] = num;ans[0] = num;cnt = 1;anscnt = 1;anssum = 0;len = num.length();rj = false;dfs(1);if(anssum==0)cout<<"error\n";else if(rj)cout<<"rejected\n";else{cout<<anssum;for(int i = 0; i < anscnt; i++)cout<<' '<<ans[i];cout<<endl;}}}

总结: 一开始直接用int数组来处理分割的数字, 结果因为多个0连在一起的原因WA了, 改用字符串处理. 这题可以不做任何剪枝, 对每个间隙分两种情况: 分割和不分割, 然后对叶子节点判断就好了, sum重复则输出rejected, 无答案则error.

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