RMQ题集+模板 hdu 2888+3486+3183 范围最小值 范围最大值

RMQ即范围最小值问题（或最大值）

先给上Sparse-Table算法 模板

void RMQ_make()
{
    int i,j;
    for(j = 1; (1<<j)<=n; ++j)
    {
        for(i = 0; i + (1<<j) - 1 < n; ++i)
            d[i][j] = max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
    }
}
int RMQ(int L, int R)
{
    int k = 0;
    while((1<<(k+1)) <= R-L+1) ++k;
    return max(d[L][k],d[R-(1<<k)+1][k]);
}

hdu 3183 有很多做法，我用贪心写的，需要RMQ的代码可以百度上搜到。

hdu 3486 枚举选择的人数m。有些人想用二分，必须注意，这个总分sum和选择的人数不是线性递增的关系。因此必须枚举。

hdu 2888 二维的RMQ，细节比较多，有点麻烦。

代码：

//HDU 3183
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#define MS(x,y) memset(x,y,sizeof(x))
#define pi acos(-1.0)
using namespace std;
void fre(){freopen("t.txt","r",stdin);}
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int MAXN = 4*(1e5)+1;
const int eps = -(1<<30);

int main()
{
    //fre();
    char n[1005];
    int m,i,j,len,cnt,flag;
    while(~scanf("%s%d",n,&m))
    {
        len = strlen(n);
        cnt = 0;
        if(m >= len){printf("0\n");continue;}
        while(cnt < m)
        {
            flag = 0;
            for(i = 0; i < len-1 ;++i)
            {
                while(n[i]==0)++i;
                j = i+1;
                while(n[j]==0)++j;
                if(n[i]>n[j]) {n[i]=0;flag = 1;cnt++;break;}
            }
            if(!flag)break;
        }
        for(i = len-1; i>=0&&cnt<m; --i,++cnt)
        {
            n[i] = 0;
        }
        i = 0;
        while(n[i]==0||n[i]=='0')i++;
        if(i >= len) {printf("0\n");continue;}
        for(; i < len; ++i)
        {
            if(n[i]) printf("%c",n[i]);
        }
        puts("");
    }
}

//HDU 3486
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#define MS(x,y) memset(x,y,sizeof(x))
#define pi acos(-1.0)
using namespace std;
void fre(){freopen("t.txt","r",stdin);}
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int MAXN = 200005;
const int eps = -(1<<30);

int n,m,d[MAXN][22];
void RMQ_make()
{
    int i,j;
    for(j = 1; (1<<j)<=n; ++j)
    {
        for(i = 0; i + (1<<j) - 1 < n; ++i)
            d[i][j] = max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
    }
}
int RMQ(int L, int R)
{
    int k = 0;
    while((1<<(k+1)) < R-L+1) ++k;
    return max(d[L][k],d[R-(1<<k)+1][k]);
}
bool solve(int x)
{
    int i,j,k = n/x;
    LL sum=0;
    for(i = 0; i < x; ++i)
    {
        sum +=RMQ(i*k,(i+1)*k-1);
        if(sum > m)break;
    }
    if(sum > m) return 1;
    else return 0;
}
int main()
{
   // fre();
    int i,j,maxx,sum;
    while(~scanf("%d%d",&n,&m))
    {
        maxx= sum=0;
        if(n<0&&m<0)break;
        for(i = 0; i < n; ++i) //input
        {
            scanf("%d",&d[i][0]);
            sum+=d[i][0];
            if(maxx < d[i][0]) maxx = d[i][0];
        }

        if(sum<=m) {printf("-1\n");continue;}//special judge

        RMQ_make();
        i = m/maxx ;//solve
        if(i==0)i++;
        while(i<n&&!solve(i))++i;
        printf("%d\n",i);
    }
}

//HDU 2888
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#define MS(x,y) memset(x,y,sizeof(x))
#define pi acos(-1.0)
using namespace std;
void fre(){freopen("t.txt","r",stdin);}
typedef long long LL;
typedef unsigned long long ULL;
const int N = 305;
const int INF = 0x3f3f3f3f;
const int M = 11;
const int eps = -(1<<30);

int row,col;
int __pow[M];
int dp[N][M][N][M];
int a[N][N];

bool check(int res , int r1 , int c1 , int r2 , int c2)
{
    if(res == a[r1][c1] || res == a[r1][c2] || res == a[r2][c1] || res == a[r2][c2]) return true;
    return false;
}

inline int max(int aa ,int bb ,int cc ,int dd)
{
    int res = -INF;
    res = aa > res ? aa : res;
    res = bb > res ? bb : res;
    res = cc > res ? cc : res;
    res = dd > res ? dd : res;
    return res;
}

void ST()
{
    int KR = (int)(log((double)row) / log(2.0));
    int KC = (int)(log((double)col) / log(2.0));
    for(int i=1; i<=row; i++)            //初始化dp[i][0][j][0]
        for(int j=1; j<=col; j++)
            dp[i][0][j][0] = a[i][j];
    for(int i=1; i<=row; i++)           //初始化dp[i][0][j][p]
        for(int pc=1; pc<=KC; pc++)
            for(int j=1; j+__pow[pc]-1<=col; j++)
            {
                int kc = j + __pow[pc-1];
                int x = dp[i][0][j][pc-1] , y = dp[i][0][kc][pc-1];
                dp[i][0][j][pc] = x > y ? x : y;
            }
    for(int j=1; j<=col; j++)           //初始化dp[i][p][j][0]
        for(int pr=1; pr<=KR; pr++)
            for(int i=1; i+__pow[pr]-1<=row; i++)
            {
                int kr = i + __pow[pr-1];
                int x = dp[i][pr-1][j][0] , y = dp[kr][pr-1][j][0];
                dp[i][pr][j][0] = x > y ? x : y;
            }
    for(int pr=1; pr<=KR; pr++)
        for(int pc=1; pc<=KC; pc++)
            for(int i=1; i+__pow[pr]-1<=row; i++)
                for(int j=1; j+__pow[pc]-1<=col; j++)
                {
                    int kr = i + __pow[pr-1];
                    int kc = j + __pow[pc-1];
                    dp[i][pr][j][pc] = max(dp[i][pr-1][j][pc-1] , dp[i][pr-1][kc][pc-1] , dp[kr][pr-1][j][pc-1] , dp[kr][pr-1][kc][pc-1]);
                }
}

int RMQ(int r1 , int c1 , int r2 , int c2)
{
    int KR = (int)(log((double)(r2-r1+1)) / log(2.0));
    int KC = (int)(log((double)(c2-c1+1)) / log(2.0));
    int kr = r2 - __pow[KR] + 1;
    int kc = c2 - __pow[KC] + 1;
    return max(dp[r1][KR][c1][KC] , dp[r1][KR][kc][KC] , dp[kr][KR][c1][KC] , dp[kr][KR][kc][KC]);
}

int main()
{
    for(int i=0; i<M; i++) __pow[i] = (1<<i);
    while(scanf("%d%d",&row,&col)!=EOF)
    {
        for(int i=1; i<=row; i++)
            for(int j=1; j<=col; j++)
                scanf("%d",&a[i][j]);
        ST();
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int ok = 0,r1,c1,r2,c2;
            scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
            int res = RMQ(r1,c1,r2,c2);
            printf("%d ",res);
            if(check(res ,r1,c1,r2,c2)) printf("yes\n");
            else                        printf("no\n");
        }
    }
    return 0;
}