poj2019cornfields【二维RMQ】

来源：互联网发布：弩配件在淘宝怎么搜编辑：程序博客网时间：2024/05/21 03:25

本来想自己写来着，憋了半天没写明白，突然意识到这玩意我有邝斌的模板啊~·~WA了n发之后才发现自己错哪了，就不能细心点，唉，比赛可咋整

/***********poj20192016.1.1927388K579MSC++1931B***********/#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define MAXN 55555#define MAXM 252int val[310][310];int dp[310][310][9][9];//最大值int dp1[300][300][9][9];int mm[310];//二进制位数减一，使用前初始化void initRMQ(int n,int m){for(int i = 1;i <= n;i++)    for(int j = 1;j <= m;j++)        dp1[i][j][0][0] =dp[i][j][0][0] = val[i][j];for(int ii = 0; ii <= mm[n]; ii++)    for(int jj = 0; jj <= mm[m]; jj++)        if(ii+jj)        for(int i = 1; i + (1<<ii) - 1 <= n;i++)            for(int j = 1; j + (1<<jj) - 1 <= m;j++)            {                if(ii)dp[i][j][ii][jj] =max(dp[i][j][ii-1][jj],dp[i+(1<<(ii-1))][j][ii-1][jj]);                else dp[i][j][ii][jj] =max(dp[i][j][ii][jj-1],dp[i][j+(1<<(jj-1))][ii][jj-1]);                if(ii)dp1[i][j][ii][jj] =min(dp1[i][j][ii-1][jj],dp1[i+(1<<(ii-1))][j][ii-1][jj]);                else dp1[i][j][ii][jj] =min(dp1[i][j][ii][jj-1],dp1[i][j+(1<<(jj-1))][ii][jj-1]);            }}//查询矩形内的最大值(x1<=x2,y1<=y2)int rmq(int x1,int y1,int x2,int y2){    int k1 = mm[x2-x1+1];    int k2 = mm[y2-y1+1];    x2 = x2 - (1<<k1) + 1;    y2 = y2 - (1<<k2) + 1;    int tmp=max(max(dp[x1][y1][k1][k2],dp[x1][y2][k1][k2]),max(dp[x2][y1][k1][k2],dp[x2][y2][k1][k2]))-min(min(dp1[x1][y1][k1][k2],dp1[x1][y2][k1][k2]),min(dp1[x2][y1][k1][k2],dp1[x2][y2][k1][k2]));    return tmp;}int main(){    //freopen("cin.txt","r",stdin);    int n,b,k;    mm[0] = -1;    for(int i = 1;i <= 253;i++)        mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];    while(~scanf("%d%d%d",&n,&b,&k))    {        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                scanf("%d",&val[i][j]);        initRMQ(n,n);        while(k--)        {            int a,se;            scanf("%d%d",&a,&se);            printf("%d\n",rmq(a,se,a+b-1,se+b-1));        }    }    return 0;}

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