洛谷P3444 [POI2006]ORK-Ploughing [枚举,贪心]

  题目传送门

ork

格式难调,体面就不放了。


  分析:

  一道偏难的贪心和枚举题。考试的时候是弃疗了。。。yyb巨佬已经讲的很详细了,推荐他的博客。这里小蒟蒻就只放代码了。

  Code:

 1 //It is made by HolseLee on 15th July 2018
 2 //Luogu.org P3444
 3 #include<bits/stdc++.h>
 4 using namespace std;
 5 const int N=2018;
 6 int n,m,K,a[N][N],ans=1e9;
 7 int sx[N][N],sy[N][N];
 8 inline int read()
 9 {
10     char ch=getchar();int num=0;bool flag=false;
11     while(ch<'0'||ch>'9'){if(ch=='-')flag=true;ch=getchar();}
12     while(ch>='0'&&ch<='9'){num=num*10+ch-'0';ch=getchar();}
13     return flag?-num:num;
14 }
15 inline int workup(int up)
16 {
17     int x1=1,x2=n,y1=1,y2=m,ret=0,ka;
18     while(x1<=x2&&y1<=y2){
19         ret++;
20         ka=sy[x2][y1]-sy[x1-1][y1];
21         if(ka<=K){++y1;continue;};
22         ka=sy[x2][y2]-sy[x1-1][y2];
23         if(ka<=K){--y2;continue;}
24         ka=sx[x1][y2]-sx[x1][y1-1];
25         if(ka<=K&&x1<up){++x1;continue;}
26         ka=sx[x2][y2]-sx[x2][y1-1];
27         if(ka<=K){--x2;continue;}
28         ret=1e9;break;
29     } 
30     return ret;
31 } 
32 inline int workle(int left)
33 {
34     int x1=1,x2=n,y1=1,y2=m,ret=0,ka;
35     while(x1<=x2&&y1<=y2){
36         ret++;
37         ka=sx[x1][y2]-sx[x1][y1-1];
38         if(ka<=K){++x1;continue;}
39         ka=sx[x2][y2]-sx[x2][y1-1];
40         if(ka<=K){--x2;continue;}
41         ka=sy[x2][y1]-sy[x1-1][y1];
42         if(ka<=K&&y1<left){++y1;continue;}
43         ka=sy[x2][y2]-sy[x1-1][y2];
44         if(ka<=K){--y2;continue;}
45         ret=1e9;break;
46     }
47     return ret;
48 }
49 int main()
50 {
51     freopen("ork.in","r",stdin);
52     freopen("ork.out","w",stdout);
53     K=read();m=read();n=read();
54     for(int i=1;i<=n;i++)
55     for(int j=1;j<=m;j++)
56     a[i][j]=read();
57     for(int i=1;i<=n;i++)
58     for(int j=1;j<=m;j++){
59     sx[i][j]=sx[i][j-1]+a[i][j];
60     sy[i][j]=sy[i-1][j]+a[i][j];}
61     for(int i=1;i<=n;i++)
62     ans=min(ans,workup(i));
63     for(int j=1;j<=m;j++)
64     ans=min(ans,workle(j));
65     printf("%d\n",ans);
66     return 0;
67 }

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转载自www.cnblogs.com/cytus/p/9314660.html