NKGOLF – SPOJ

April 29, 2016 by Leave a Comment

Đề bài: http://vn.spoj.com/problems/NKGOLF/

Thuật toán:

  • (đang cập nhập)

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <utility>
#include <sstream>
using namespace std;
 
#define FOR(i,a,b) for(int i = (a); i <= (b); i++)
#define DOW(i,a,b) for(int i = (a); i >= (b); i--)
#define RESET(c,val) memset(c,val,sizeof(c))
#define sz(a) int(a.size())
#define pb push_back
 
typedef long long ll;
typedef unsigned long long ull;
 
/*---------------------------*/
 
const int maxn = 1e3 + 2;
int m, n, res, tt;
ll a[maxn][maxn];
bool b[maxn][maxn];
 
void input() {
    scanf("%d %d", &m, &n);
    RESET(a,0);
    FOR(i,1,m)
    FOR(j,1,n)
    scanf("%lld", &a[i][j]);
}
 
void solve() {
    FOR(i,1,m-1)
    FOR(j,1,n-1)
    b[i][j]=((a[i][j]<=a[i+1][j]) && (a[i][j]<=a[i][j+1])
             && (a[i+1][j]<=a[i+1][j+1]) && (a[i][j+1]<=a[i+1][j+1]));
 
    res=1;
    // search row
    FOR(i,1,m) {
        tt=1;
        FOR(j,2,n)
        if ( a[i][j]>=a[i][j-1] ) {
            tt++;
            res=max(res,tt);
        } else tt=1;
    }
 
    // search col
    FOR(j,1,n) {
        tt=1;
        FOR(i,2,m)
        if ( a[i][j]>=a[i-1][j] ) {
            tt++;
            res=max(res,tt);
        } else tt=1;
    }
 
    int h[maxn], l[maxn], r[maxn];
    RESET(h,0);
    RESET(l,0);
    RESET(r,0);
    FOR(i,1,m-1) {
        FOR(j,1,n-1)
        if ( b[i][j] ) h[j]++;
        else h[j]=0;
 
        //deque
        int d[maxn];
        int top=0;
        d[0]=0;
        FOR(j,1,n-1) {
            while(top && h[j]<=h[d[top]])
                r[d[top--]]=j-1;
            l[j]=d[top]+1;
            d[++top]=j;
        }
        while(top) r[d[top--]]=n-1;
 
        FOR(j,1,n-1)
        if ( h[j]>0 ) {
            res=max(res,(h[j]+1)*(r[j]-l[j]+2));
        }
    }
}
 
void output() {
    printf("%d", res);
}
 
int main() {
    input();
    solve();
    output();
 
    return 0;
}
Chủ đề: SPOJ, Thuật toán Mức độ: ,

CREC01 – SPOJ

April 3, 2016 by Leave a Comment

đề bài: http://vn.spoj.com/problems/CREC01/

Thuật toán:

  • Bài này ta sử dụng kỹ thuật deque. Thuật toán cũng không có gì phức tạp. Bạn có thể đọc code bên dưới để hiểu thêm

Code:

Pascal:

const   fi      ='';
        fo      ='';
        oo      =1000;
 
var     a       :array[0..oo,0..oo] of 0..1;
        l,h,dem:array[0..oo+1] of int64;
        m, n    :longint;
 
procedure Optimize;
var     i, j    :longint;
        ans     :int64;
begin
        fillchar(h, sizeof(h),0);
        ans:=0;
        for i:=1 to m do
                begin
                        for j:=1 to n do
                                begin
                                        if a[i,j]=1 then begin
                                        l[j]:=j;
                                        h[j]:=h[j]+1;
                                        while h[l[j]-1]>h[j] do
                                                l[j]:=l[l[j]-1];
                                        dem[j]:=h[j]*(j-l[j]+1)+dem[l[j]-1];
                                        ans:=ans+dem[j];
                                        //writeln(ans,'==');
                                        //if (i=3) and (j=2) then writeln(dem[1]);
                                        end
                                        else begin
                                                dem[j]:=0;
                                                l[j]:=0;
                                                h[j]:=0;
                                        end;
 
                                end;
 
                end;
        writeln(ans);
end;
 
procedure run;
var     i, j, tg    :longint;
        s       :ansistring;
begin
        assign(input,fi);assign(output,fo);
        reset(input);rewrite(output);
        readln(m, n);
        for i:=1 to m do
                begin
                        readln(s);
                        for j:=1 to n do
                                        val(s[j],a[i,j]);
                end;
        Optimize;
        close(input);close(output);
 
end;
 
begin
        run;
end.

C++:

using namespace std;
#include<bits/stdc++.h>
 
#define BG begin()
#define ED end()
#define st first
#define nd second
#define PB push_back
#define PF push_front
#define FOR(i,a,b) for (long long i=a;i<b;i++)
#define FORE(i,a,b) for (long long i=a;i<=b;i++)
#define FORD(i,a,b) for (long long i=a;i>=b; i--)
#define ri(n)({
    int neg=0;
    n=0;
    char ch;
    for(ch=getchar(); ch<'0' || ch>'9'; ch=getchar()) if (ch=='-') neg=1-neg;
    n=ch-48;
    for(ch=getchar(); ch>='0' && ch<='9'; ch=getchar()) n=(n<<3)+(n<<1)+ch-48;
    if (neg) n=-n;
})
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> II;
typedef pair<ll,ll> LL;
const ll INF=1000000000+7;
const double esp=1e-13;
const double pi=3.141592653589;
 
 
ll m, n, a[1001][1001], H[1001], dem[1001], L[1001];
 
int main()
{
    //freopen("CREC01.inp", "r", stdin);
    //freopen("CREC01.out", "w", stdout);
    cin>>m>>n;
    char ch;
    FORE(i, 1, m)
        FORE(j, 1, n) {
            cin>>ch;
            a[i][j] = ch - '0';
        }
    memset(H, 0, sizeof(H));
    memset(L, 0, sizeof(L));
    memset(dem, 0, sizeof(dem));
    long long ans = 0;
    FORE(i, 1, m)
        FORE(j, 1, n) {
            //cout<<i<<" "<<j<<" "<<a[i][j]<<endl;
            if (a[i][j] == 1) {
                H[j]++;
                int k = j;
                while ( (k > 1) && (a[i][k-1] == 1) && (H[ k - 1 ] >= H[j]) ) k = L[k - 1];
                L[j] = k;
                dem[j] = H[j]*(j - L[j] + 1) + dem[L[j] - 1];
                ans += dem[j];
                //cout<<i<<"=="<<j<<"=="<<"=="<<H[j]<<"=="<<ans<<endl;
            }
            else
            {
                H[j] = 0;
                dem[j]= 0;
                L[j] = 0;
            }
        }
    cout<<ans;
}
Chủ đề: SPOJ Mức độ: