您的位置:首页 > 其它

CodeForces 15D Map 单调队列优化

2014-10-30 10:52 375 查看
两次单调队列求出每个子矩阵的最小值,区间减法求出每个子矩阵的和,然后丢到优先队列里跑出来就好了。

写锉了,加了读入挂才过。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007

/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
char c;
while(!d);
x=c-'0';
while(d)p;
return x;
}
template<class T> inline T& RDD(T &x)
{
char c;
while(g,c!='-'&&!isdigit(c));
if (c=='-')
{
x='0'-g;
while(d)n;
}
else
{
x=c-'0';
while(d)p;
}
return x;
}
inline double& RF(double &x)      //scanf("%lf", &x);
{
char c;
while(g,c!='-'&&c!='.'&&!isdigit(c));
if(c=='-')if(g=='.')
{
x=0;
double l=1;
while(d)nn;
x*=l;
}
else
{
x='0'-c;
while(d)n;
if(c=='.')
{
double l=1;
while(d)nn;
x*=l;
}
}
else if(c=='.')
{
x=0;
double l=1;
while(d)pp;
x*=l;
}
else
{
x=c-'0';
while(d)p;
if(c=='.')
{
double l=1;
while(d)pp;
x*=l;
}
}
return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;

const int BLOCK = sqrt(1000.0);

LL Map[1010][1010];

bool mark[1010][1010];

LL ans[1010][1010] = {0};

LL Min[1010][1010];

struct N
{
int x,y;
LL val;
bool operator < (const N &a) const
{
if(val != a.val)
return a.val < val;
if(x != a.x)
return a.x < x;
return a.y < y;
}
} tmp;

int x[1000100],y[1000100];
LL val[1000100];

priority_queue<N> q;

int que[1010];

void CalMin(int n,int m,int a,int b)
{
int i,j,S,E;

for(i = 1; i <= n; ++i)
{
S = 0,E = 0;
for(j = m; j >= 1; --j)
{
while(E > S && Map[i][que[E-1]] > Map[i][j])
E--;
que[E++] = j;
while(S < E && que[S] > j+b-1)
S++;

Min[i][j] = Map[i][que[S]];
}
}

for(j = 1; j <= m; ++j)
{
S = 0,E = 0;
for(i = n; i >= 1; --i)
{
while(E > S && Min[que[E-1]][j] > Min[i][j])
E--;
que[E++] = i;
while(S < E && que[S] > i+a-1)
S++;
Map[i][j] = Min[que[S]][j];
}
}
}

int main()
{
int i,j,n,m,a,b;
RD(n);
RD(m);
RD(a);
RD(b);
// scanf("%d %d %d %d",&n,&m,&a,&b);

for(i = 1; i <= n; ++i)
for(j = 1; j <= m; ++j)
RD(Map[i][j]);//scanf("%I64d",&Map[i][j]);

for(i = n; i >= 1; --i)
for(j = m; j >= 1; --j)
ans[i][j] = Map[i][j] + ans[i+1][j] + ans[i][j+1] - ans[i+1][j+1];

int N = n-a+1,M = m-b+1;

for(i = 1; i <= N; ++i)
for(j = 1; j <= M; ++j)
ans[i][j] = ans[i][j] - ans[i+a][j] - ans[i][j+b] + ans[i+a][j+b];

CalMin(n,m,a,b);

//    for(i = 1;i <= n; ++i)
//    {
//        for(j = 1;j <= m; ++j)
//            printf("%3I64d ",ans[i][j]);
//        puts("");
//    }
//
//    puts("");
//
//    for(i = 1;i <= n; ++i)
//    {
//        for(j = 1;j <= m; ++j)
//            printf("%3I64d ",Min[i][j]);
//        puts("");
//    }

for(i = N; i >= 1; --i)
for(j = M; j >= 1; --j)
q.push( {i,j,ans[i][j]-Map[i][j]*a*b});
memset(mark,false,sizeof(mark));

int L,R,T,B,Top = 0;

while(q.empty() == false)
{
tmp = q.top();
q.pop();

if(mark[tmp.x][tmp.y])
continue;

T = max(1,tmp.x-a+1),B = min(n,tmp.x+a-1);
L = max(1,tmp.y-b+1),R = min(m,tmp.y+b-1);

for(i = T; i <= B; ++i)
for(j = L; j <= R; ++j)
mark[i][j] = true;
x[Top] = tmp.x,y[Top] = tmp.y,val[Top] = tmp.val,Top++;
}

printf("%d\n",Top);

for(i = 0; i < Top; ++i)
printf("%d %d %I64d\n",x[i],y[i],val[i]);

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: