One hundred layer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1257 Accepted Submission(s): 467
Problem Description
Now there is a game called the new man down 100th floor. The rules of this game is:
1. At first you are at the 1st floor. And the floor moves up. Of course you can choose which part you will stay in the first time.
2. Every floor is divided into M parts. You can only walk in a direction (left or right). And you can jump to the next floor in any part, however if you are now in part “y”, you can only jump to part “y” in the next floor! (1<=y<=M)
3. There are jags on the ceils, so you can only move at most T parts each floor.
4. Each part has a score. And the score is the sum of the parts’ score sum you passed by.
Now we want to know after you get the 100th floor, what’s the highest score you can get.
1. At first you are at the 1st floor. And the floor moves up. Of course you can choose which part you will stay in the first time.
2. Every floor is divided into M parts. You can only walk in a direction (left or right). And you can jump to the next floor in any part, however if you are now in part “y”, you can only jump to part “y” in the next floor! (1<=y<=M)
3. There are jags on the ceils, so you can only move at most T parts each floor.
4. Each part has a score. And the score is the sum of the parts’ score sum you passed by.
Now we want to know after you get the 100th floor, what’s the highest score you can get.
Input
The first line of each case has four integer N, M, X, T(1<=N<=100, 1<=M<=10000, 1<=X， T<=M). N indicates the number of layers; M indicates the number of parts. At first you are in the X-th part. You can move at most T parts in every floor in
only one direction.
Followed N lines, each line has M integers indicating the score. (-500<=score<=500)
Followed N lines, each line has M integers indicating the score. (-500<=score<=500)
Output
Output the highest score you can get.
Sample Input
3 3 2 1 7 8 1 4 5 6 1 2 3
Sample Output
29
Source
Recommend
zhuyuanchen520
dp[i][j] = dp[i-1][k] + dis(k,j);
dp[i][j] = dp[i-1][1] + dis(1,j);
dp[i][j] = dp[i-1][2] + dis(2,j);
...
转换为
dp[i][j] = dp[i-1][1] + dis(1,j);
dp[i][j] = dp[i-1][1] + dis(2,j)+[dis(2,j)-dis(1,j)];
...
这样只要比较dis(1,j)和dis(2,j)+[dis(2,j)-dis(1,j)] ...
在距离限制内维护一个单调队列可以O(1)取到最值
--------
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) {
REP(i, n){
REP(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
#define Display_1(A, n, m) {
REP_1(i, n){
REP_1(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
/**
f[i][j]=max(f[i-1][k]+sum(k,j))
=max(f[i-1][k]+sum(1,j)-sum(1,k))
=max(f[i-1][k]-sum(1,k))+sum(1,j)
**/
const int maxn=111;
const int maxm=11111;
int n,m,X,T;
int score[maxn][maxm];
int f[maxn][maxm];
int que[maxm],les[maxm];
int main()
{
int head,tail;
int sum,tmp;
while (cin>>n>>m>>X>>T)
{
REP_1(i,n) REP_1(j,m)
cin>>score[i][j];
FOR_1(i,0,m) f[0][i]=-INF;
f[0][X]=0;
REP_1(i,n)
{
f[i][1]=f[i-1][1]+score[i][1];
head=tail=0;
les[0]=1;
que[0]=f[i-1][1];
sum=score[i][1];
FOR_1(j,2,m)
{
while (head<=tail&&j-les[head]>T) head++;
tmp=f[i-1][j]-sum;
while (head<=tail&&que[tail]<=tmp) tail--;
sum+=score[i][j];
tail++;
les[tail]=j;
que[tail]=tmp;
f[i][j]=que[head]+sum;
}
head=tail=0;
les[0]=m;
f[i][m]=max(f[i][m],f[i-1][m]+score[i][m]);
que[0]=f[i-1][m];
sum=score[i][m];
DWN_1(j,m-1,1)
{
while (head<=tail&&les[head]-j>T) head++;
tmp=f[i-1][j]-sum;
while (head<=tail&&que[tail]<=tmp) tail--;
sum+=score[i][j];
tail++;
les[tail]=j;
que[tail]=tmp;
f[i][j]=max(f[i][j],que[head]+sum);
}
}
int ans=-INF;
REP_1(i,m)
{
ans=max(ans,f[n][i]);
}
cout<<ans<<endl;
}
return 0;
}