RMRC2016 G:Flow Shop (DP)

 Flow Shop

时间限制: 6 Sec  内存限制: 128 MB
提交: 16  解决: 15
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题目描述

harvest grain). All swathers go through the same basic stages in their construction: for example they all need to have a cutting bar, a grain belt, and a reel fitted. However, these components can be customized based on the buyer’s needs, so these various stages may take different amounts of time between different swathers.
N swathers have been ordered and there are M stages in the manufacturing process. The swathers will each go through the same sequence of stages.
In particular, the processing occurs as follows: For each swather i and each stage j, it takes Pij units of time to complete stage j for swather i. The workers at each stage may only work on one swather at a time. At the start of the day all swather orders are ready to be processed by the first stage. At any point in the process, if the workers at stage j are idle and there are swathers waiting to be processed at this stage then the workers will pick the swather that has the lowest label (they are labelled from 1 to N). Note that the work on a stage j can only be started after the work on the stage j-1 is completed.
Determine the time each swather is completed.

输入

There is only one test case in each file. It begins with a single line containing N andM (1≤N,M≤1000),the number of swathers and stages (respectively). Following this are N lines, each withM integers. The j'th integer of the i’th line is Pij , giving the amount of time it will take for the workers at stage j to complete swather i (1≤Pij≤106).

输出

Output a single line containing N integers T1 T2...Tn with a single space between consecutive integers.
These should be such that stage M for swather i is completed at time Ti.

样例输入

2 3

1 2 3

3 2 1

样例输出

6 7

这道题  是严格按照 顺序 来的  所以说 也不能算是 dp

如果 按照dp  讲的 话  枚举列  对于 每行  判断 dp[j] 与 dp[j-1] 的 关系

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n)   scanf("%I64d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)#define Pr(n)     printf("%d\n",n)using namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};int maps[1100][1100];int dp[1200];int main(){    int n,m;    while(~S2(n,m))    {        mem(maps,0);        mem(dp,0);        int sum=0;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                S1(maps[i][j]);        int ans=0;        for(int i=1;i<=n;i++)            dp[i]+=maps[i][1];        for(int i=2;i<=m;i++)//列            for(int j=1;j<=n;j++)// 行            {                if(j==1)                    dp[j]+=maps[j][i];                else                {                    if(dp[j]>dp[j-1])                        dp[j]+=maps[j][i];                    else                        dp[j]=dp[j-1]+maps[j][i];                }            }        for(int i=1;i<=n;i++)        {            if(i==1)                printf("%d",dp[i]);            else                printf(" %d",dp[i]);        }        cout<<endl;    }    return 0;}

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