poj 1017

来源：互联网发布：淘宝怎么批量修改价格编辑：程序博客网时间：2024/05/16 14:43
模拟题，我们按照先放体积大的，再放体积小的贪心思想来放，就能够使得盒子总数最少，具体看注释。
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m;#define M 110#define N 1000010#define Mod 258280327#define p(x,y) make_pair(x,y)const int MAX_len=550;int res[50010];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    int num[10];    while(scanf("%d%d%d%d%d%d",&num[1],&num[2],&num[3],&num[4],&num[5],&num[6])!=EOF&&num[1]+num[2]+num[3]+num[4]+num[5]+num[6]){        int box=0;        if(num[6]){     //如果体积是6*6的话，那就一个产品对应一个盒子            box += num[6];            num[6]=0;        }        if(num[5]){            box += num[5];  //如果体积是5*5的话，那一个产品对应一个盒子            num[1] = Max(num[1]-num[5]*11,0);   //剩下的还可以装num[5]*11个1*1的盒子            num[5]=0;        }        if(num[4]){            box += num[4];  //如果体积是4*4的话，那一个产品对应一个盒子            int tot = num[4]*20;    //这是num[4]个的盒子剩下的总体积            if(num[2]){                int num2 = tot/4;   //可以放num2个2*2的盒子                if(num2>num[2]){                    if(num[1]){ //还能放下(num2-num[2]*4个1*1的盒子                        int num1 = num2-num[2];                        num[1] = Max(num[1]-num1*4,0);                    }                    num[2] = 0;                }else{                    num[2] -= num2;                }            }            num[4]=0;        }        if(num[3]){            box += (num[3]/4);  //如果体积是3*3的话，那么4个3*3对应一个盒子            if(num[3]%4)                box++;            int num3 = num[3]%4;    //最后一个放3*3的盒子里面有几个3*3            if(num3 == 1){  //如果只放了一个3*3，那么可以至多放5个2*2的盒子以及7个1*1的盒子                if(num[2]>=5){                    num[2] -= 5;                    if(num[1]){                        num[1] = Max(num[1]-7,0);                    }                }else{                    if(num[1]){                        num[2] = 5-num[2];                        num[1] = Max(num[1]-7-num[2]*4,0);                    }                    num[2] = 0;                }            }if(num3 == 2){ //如果放了2个3*3的盒子，那么至多可以放3个2*2的盒子和6个1*1的盒子                if(num[2]>=3){                    num[2] -= 3;                    if(num[1]){                        num[1] = Max(num[1]-6,0);                    }                }else{                    if(num[1]){                        num[2] = 3-num[2];                        num[1] = Max(num[1]-6-num[2]*4,0);                    }                    num[2] = 0;                }            }if(num3 == 3){     //如果放了3个3*3的盒子，那么至多还可以放1个2*2的盒子和5个1*1的盒子                if(num[2]){                    num[2] -= 1;                    if(num[1]){                        num[1] = Max(num[1]-5,0);                    }                }else{                    if(num[1]){                        num[2] = 1-num[2];                        num[1] = Max(num[1]-5-num[2]*4,0);                    }                    num[2] = 0;                }            }            num[3] = 0;        }        if(num[2]){     //剩下的还有2*2的盒子            box += num[2]/9;            if(num[2]%9)                box++;            int num2 = num[2]%9;            if(num2){                num2 = 9-num2;                num[1] = Max(num[1]-num2*4,0);                num[2]=0;            }        }        if(num[1]){ //剩下的还有1*1的盒子            box += num[1]/36;            if(num[1]%36)                box++;        }        printf("%d\n",box);    }    return 0;}
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