关于二维矩阵的最大最小值的询问

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今天考了这样子的一道题目，考试的时候yy出了二维RMQ的做法，调了半天，考完后，想了想二维线段树的做法,感觉很好想的，只是把一维线段树改一改。

二维RMQ

处理出dp[i][j][l][r]表示i往前2l，j往前2r的矩阵的最大值和最小值，首先是dp[i][j][0][0]=原数值，然后每次枚举l，r然后枚举i，j就好了。像一维前缀和那样。询问的时候变成4个矩形就好了。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;typedef long long LL;#define REP(i,a,b) for(register int i = (a),i##_end_ = (b); i <= i##_end_; ++i)#define DREP(i,a,b) for(register int i = (a),i##_end_ = (b); i >= i##_end_; --i)#define mem(a,b) memset(a,b,sizeof(a))template<typename T> inline bool chkmin(T &a,const T &b) { return a > b ? a = b , 1 : 0;}template<typename T> inline bool chkmax(T &a,const T &b) { return a < b ? a = b , 1 : 0;}int read(){    register int f = 1,s = 0;char c = getchar();    while(!isdigit(c)) { if(c == '-')f = -1;c = getchar();}    while(isdigit(c)) { s = s * 10 + c - '0';c = getchar();}    return f * s;}const int maxn = 510;int n,m;int a[maxn][maxn],dp[maxn][maxn][10][10],s[maxn][maxn];int f[maxn][maxn][10][10];char p[20];void init(){    for (int i = 0; (1 << i) <= n; i++)    {        for (int j = 0; (1 << j) <= m; j++)        {            if (i == 0 && j == 0) continue;            for (int row = 1; row + (1 << i) - 1 <= n; row++)                for (int col = 1; col + (1 << j) - 1 <= m; col++)                {                    if (i == 0)         dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]);                    else if (j == 0)          dp[row][col][i][j] = max(dp[row][col][i - 1][j], dp[row + (1 << (i - 1))][col][i - 1][j]);                    else         dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]);                }        }    }    for (int i = 0; (1 << i) <= n; i++)    {        for (int j = 0; (1 << j) <= m; j++)        {            if (i == 0 && j == 0) continue;            for (int row = 1; row + (1 << i) - 1 <= n; row++)                for (int col = 1; col + (1 << j) - 1 <= m; col++)                {                    if (i == 0)         f[row][col][i][j] = min(f[row][col][i][j - 1], f[row][col + (1 << (j - 1))][i][j - 1]);                    else if (j == 0)          f[row][col][i][j] = min(f[row][col][i - 1][j], f[row + (1 << (i - 1))][col][i - 1][j]);                    else         f[row][col][i][j] = min(f[row][col][i][j - 1], f[row][col + (1 << (j - 1))][i][j - 1]);                }        }    }}int Query_Max(int x1, int y1, int x2, int y2) {    int kx = 0, ky = 0;    while (x1 + (1 << (1 + kx)) - 1 <= x2) kx++;    while (y1 + (1 << (1 + ky)) - 1 <= y2) ky++;    int m1 = dp[x1][y1][kx][ky];    int m2 = dp[x2 - (1 << kx) + 1][y1][kx][ky];    int m3 = dp[x1][y2 - (1 << ky) + 1][kx][ky];    int m4 = dp[x2 - (1 << kx) + 1][y2 - (1 << ky) + 1][kx][ky];    return max(max(m1, m2), max(m3, m4));}int Query_Min(int x1, int y1, int x2, int y2) {    int kx = 0, ky = 0;    while (x1 + (1 << (1 + kx)) - 1 <= x2) kx++;    while (y1 + (1 << (1 + ky)) - 1 <= y2) ky++;    int m1 = f[x1][y1][kx][ky];    int m2 = f[x2 - (1 << kx) + 1][y1][kx][ky];    int m3 = f[x1][y2 - (1 << ky) + 1][kx][ky];    int m4 = f[x2 - (1 << kx) + 1][y2 - (1 << ky) + 1][kx][ky];    return min(min(m1, m2), min(m3, m4));}int Sum(int x1,int y1,int x2,int y2){    return s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1];}int main(){    freopen("phalanx.in","r",stdin);    freopen("phalanx.out","w",stdout);    n = read(),m = read();    REP(i,1,n)REP(j,1,m)    {        a[i][j] = read();        dp[i][j][0][0] = f[i][j][0][0] = a[i][j];        s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + a[i][j];    }    init();    int T = read();    while(T--)    {        cin>>p;        int x1 = read()+1,y1 = read()+1,x2 = read()+1,y2 = read()+1;        if(p[0] == 'S')printf("%d\n",Sum(x1,y1,x2,y2));        else if(p[1] == 'A')printf("%d\n",Query_Max(x1,y1,x2,y2));        else printf("%d\n",Query_Min(x1,y1,x2,y2));    }    return 0;}

二维线段树

其实跟一维线段树差不多，我们首先在每一行维护一个线段树，然后再在列上维护一个线段树，然后每次合并就好了。询问的时候先查对应的行，然后就是，递归找。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;typedef long long LL;#define REP(i,a,b) for(register int i = (a),i##_end_ = (b); i <= i##_end_; ++i)#define DREP(i,a,b) for(register int i = (a),i##_end_ = (b); i >= i##_end_; --i)#define mem(a,b) memset(a,b,sizeof(a))template<typename T> inline bool chkmin(T &a,const T &b) { return a > b ? a = b , 1 : 0;}template<typename T> inline bool chkmax(T &a,const T &b) { return a < b ? a = b , 1 : 0;}int read(){    register int f = 1,s = 0;char c = getchar();    while(!isdigit(c)) { if(c == '-')f = -1;c = getchar();}    while(isdigit(c)) { s = s * 10 + c - '0';c = getchar();}    return f * s;}const int maxn = 810;int n,m;int a[maxn][maxn];int s[maxn][maxn];int Max[maxn<<2][maxn<<2],Min[maxn<<2][maxn<<2];//[]第一维表示第一颗线段树，[]第二维表示第二棵线段树的根节点编号。void Build(int H,int row,int h,int l,int r){//递归建第二维。    if(l == r)    {        Max[H][h] = a[row][l];        Min[H][h] = a[row][l];        return ;    }    int mid = (l + r) >> 1;    Build(H,row,h<<1,l,mid);Build(H,row,h<<1|1,mid+1,r);    Max[H][h] = max(Max[H][h<<1],Max[H][h<<1|1]);    Min[H][h] = min(Min[H][h<<1],Min[H][h<<1|1]);}void merge(int H,int h1,int h2,int h,int l,int r){//合并两个儿子，把第二维合并。    if(l == r)    {        Max[H][h] = max(Max[h1][h],Max[h2][h]);        Min[H][h] = min(Min[h1][h],Min[h2][h]);        return ;    }    int mid = (l + r) >> 1;    merge(H,h1,h2,h<<1,l,mid);    merge(H,h1,h2,h<<1|1,mid+1,r);    Max[H][h] = max(Max[h1][h],Max[h2][h]);    Min[H][h] = min(Min[h1][h],Min[h2][h]);}void build(int h,int l,int r){//普通线段树建树，到叶子节点的时候再递归建树。回溯的时候，merge两个儿子。    if(l == r)    {        Build(h,l,1,1,m);        return ;    }    int mid = (l + r) >> 1;    build(h<<1,l,mid);build(h<<1|1,mid+1,r);    merge(h,h<<1,h<<1|1,1,1,m);}//递归询问.int query_max(int H,int h,int l,int r,int q,int w){    if(l == q && r == w)return Max[H][h];    int mid = (l + r) >> 1;    if(w <= mid)return query_max(H,h<<1,l,mid,q,w);    else if(q > mid)return query_max(H,h<<1|1,mid+1,r,q,w);    else return max(query_max(H,h<<1,l,mid,q,mid),query_max(H,h<<1|1,mid+1,r,mid+1,w));}int Query_max(int h,int l,int r,int q,int w,int ll,int rr){    if(l == q && r == w)return query_max(h,1,1,m,ll,rr);    int mid = (l + r) >> 1;    if(w <= mid)return Query_max(h<<1,l,mid,q,w,ll,rr);    else if(q > mid)return Query_max(h<<1|1,mid+1,r,q,w,ll,rr);    else return max(Query_max(h<<1,l,mid,q,mid,ll,rr),Query_max(h<<1|1,mid+1,r,mid+1,w,ll,rr));}int query_min(int H,int h,int l,int r,int q,int w){    if(l == q && r == w)return Min[H][h];    int mid = (l + r) >> 1;    if(w <= mid)return query_min(H,h<<1,l,mid,q,w);    else if(q > mid)return query_min(H,h<<1|1,mid+1,r,q,w);    else return min(query_min(H,h<<1,l,mid,q,mid),query_min(H,h<<1|1,mid+1,r,mid+1,w));}int Query_min(int h,int l,int r,int q,int w,int ll,int rr){    if(l == q && r == w)return query_min(h,1,1,m,ll,rr);    int mid = (l + r) >> 1;    if(w <= mid)return Query_min(h<<1,l,mid,q,w,ll,rr);    else if(q > mid)return Query_min(h<<1|1,mid+1,r,q,w,ll,rr);    else return min(Query_min(h<<1,l,mid,q,mid,ll,rr),Query_min(h<<1|1,mid+1,r,mid+1,w,ll,rr));}int Q;char ss[20];int Sum(int x1,int y1,int x2,int y2){    return s[x2][y2] - s[x2][y1-1] - s[x1-1][y2] + s[x1-1][y1-1];}int main(){    freopen("phalanx.in","r",stdin);    freopen("phalanx.out","w",stdout);    n = read(),m = read();    REP(i,1,n) REP(j,1,m)    {        a[i][j] = read();        s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + a[i][j];    }    build(1,1,n);       Q = read();    while(Q--)    {        cin>>ss;        int x1 = read()+1,y1 = read()+1,x2 = read()+1,y2 = read()+1;        if(ss[0] == 'S')printf("%d\n",Sum(x1,y1,x2,y2));        else if(ss[1] == 'A')printf("%d\n",Query_max(1,1,n,x1,x2,y1,y2));        else printf("%d\n",Query_min(1,1,n,x1,x2,y1,y2));    }    return 0;}

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