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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 1785 treap 或 RMQ

POJ 1785 treap 或 RMQ

編輯:C++入門知識

[cpp]
#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <string>  
#include <cstdio>  
#include <cmath>  
#include <queue>  
#include <map>  
#include <set>  
#define eps 1e-5  
#define MAXN 55555  
#define MAXM 5555  
#define INF 100000007  
using namespace std; 
struct node 

    int w, fa, lch, rch; 
    char s[111]; 
    bool operator <(const node& a)const 
    { 
        return strcmp(s, a.s) <= 0; 
    } 
}p[MAXN]; 
int n; 
void insert(int i) 

    int j = i - 1; 
    while(p[j].w < p[i].w) j = p[j].fa; 
    p[i].lch = p[j].rch; 
    p[j].rch = i; 
    p[i].fa = j; 

void solve(int rt) 

    if(rt == 0) return; 
    printf("("); 
    solve(p[rt].lch); 
    printf("%s/%d", p[rt].s, p[rt].w); 
    solve(p[rt].rch); 
    printf(")"); 

int main() 

    char s[111], tmp[111]; 
    while(scanf("%d", &n) != EOF && n) 
    { 
        for(int i = 1; i <= n; i++) 
        { 
            scanf("%s", s); 
            int j; 
            for(j = 0; s[j]; j++) 
                if(s[j] == '/') break; 
                    else tmp[j] = s[j]; 
            tmp[j++] = '\0'; 
            int num = 0; 
            while(s[j]) {num = num * 10 + s[j] - '0'; j++;} 
            strcpy(p[i].s, tmp); 
            p[i].w = num; 
            p[i].lch = p[i].rch = 0; 
            p[i].fa = 0; 
        } 
        sort(p + 1, p + n + 1); 
        p[0].w = INF; 
        p[0].lch = p[0].rch = p[0].fa = 0; 
        for(int i = 1; i <= n; i++) insert(i); 
        solve(p[0].rch); 
        printf("\n"); 
    } 
    return 0; 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 55555
#define MAXM 5555
#define INF 100000007
using namespace std;
struct node
{
    int w, fa, lch, rch;
    char s[111];
    bool operator <(const node& a)const
    {
        return strcmp(s, a.s) <= 0;
    }
}p[MAXN];
int n;
void insert(int i)
{
    int j = i - 1;
    while(p[j].w < p[i].w) j = p[j].fa;
    p[i].lch = p[j].rch;
    p[j].rch = i;
    p[i].fa = j;
}
void solve(int rt)
{
    if(rt == 0) return;
    printf("(");
    solve(p[rt].lch);
    printf("%s/%d", p[rt].s, p[rt].w);
    solve(p[rt].rch);
    printf(")");
}
int main()
{
    char s[111], tmp[111];
    while(scanf("%d", &n) != EOF && n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%s", s);
            int j;
            for(j = 0; s[j]; j++)
                if(s[j] == '/') break;
                    else tmp[j] = s[j];
            tmp[j++] = '\0';
            int num = 0;
            while(s[j]) {num = num * 10 + s[j] - '0'; j++;}
            strcpy(p[i].s, tmp);
            p[i].w = num;
            p[i].lch = p[i].rch = 0;
            p[i].fa = 0;
        }
        sort(p + 1, p + n + 1);
        p[0].w = INF;
        p[0].lch = p[0].rch = p[0].fa = 0;
        for(int i = 1; i <= n; i++) insert(i);
        solve(p[0].rch);
        printf("\n");
    }
    return 0;
}

然後還可以使用RMQ來搞

還是首先要排序

每次遞歸處理一個區間,找出有最大第二關鍵字的那個做根。他左邊的就是他的左子樹,右邊的是右子樹


[cpp]
#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <string>  
#include <cstdio>  
#include <cmath>  
#include <queue>  
#include <map>  
#include <set>  
#define eps 1e-5  
#define MAXN 55555  
#define MAXM 5555  
#define INF 100000007  
using namespace std; 
struct node 

    int w; 
    char s[111]; 
    bool operator <(const node& a)const 
    { 
        return strcmp(s, a.s) <= 0; 
    } 
}p[MAXN]; 
int mi[MAXN][17], mx[MAXN][17]; 
int n; 
void rmqinit() 

    for(int i = 1; i <= n; i++) mi[i][0] = mx[i][0] = i; 
    int m = (int)(log(n * 1.0) / log(2.0)); 
    for(int i = 1; i <= m; i++) 
        for(int j = 1; j <= n; j++) 
        { 
            mx[j][i] = mx[j][i - 1]; 
            mi[j][i] = mi[j][i - 1]; 
            int k = j + (1 << (i - 1)); 
            if(k <= n) 
            { 
                if(p[mx[j][i]].w < p[mx[k][i - 1]].w) mx[j][i] = mx[k][i - 1]; 
                if(p[mi[j][i]].w > p[mi[k][i - 1]].w) mi[j][i] = mi[k][i - 1]; 
            } 
        } 

int rmqmax(int l, int r) 

    int m = (int)(log((r - l + 1) * 1.0) / log(2.0)); 
    if(p[mx[l][m]].w < p[mx[r - (1 << m) + 1][m]].w) return mx[r - (1 << m) + 1][m]; 
    else return mx[l][m]; 

void solve(int s, int t) 

    if(s > t) return; 
    int pos = rmqmax(s, t); 
    printf("("); 
    solve(s, pos - 1); 
    printf("%s/%d", p[pos].s, p[pos].w); 
    solve(pos + 1, t); 
    printf(")"); 

int main() 

    char s[111], tmp[111]; 
    while(scanf("%d", &n) != EOF && n) 
    { 
        for(int i = 1; i <= n; i++) 
        { 
            scanf("%s", s); 
            int j; 
            for(j = 0; s[j]; j++) 
                if(s[j] == '/') break; 
                    else tmp[j] = s[j]; 
            tmp[j++] = '\0'; 
            int num = 0; 
            while(s[j]) {num = num * 10 + s[j] - '0'; j++;} 
            strcpy(p[i].s, tmp); 
            p[i].w = num; 
        } 
        sort(p + 1, p + n + 1); 
        rmqinit(); 
        solve(1, n); 
        printf("\n"); 
    } 
    return 0; 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 55555
#define MAXM 5555
#define INF 100000007
using namespace std;
struct node
{
    int w;
    char s[111];
    bool operator <(const node& a)const
    {
        return strcmp(s, a.s) <= 0;
    }
}p[MAXN];
int mi[MAXN][17], mx[MAXN][17];
int n;
void rmqinit()
{
    for(int i = 1; i <= n; i++) mi[i][0] = mx[i][0] = i;
    int m = (int)(log(n * 1.0) / log(2.0));
    for(int i = 1; i <= m; i++)
        for(int j = 1; j <= n; j++)
        {
            mx[j][i] = mx[j][i - 1];
            mi[j][i] = mi[j][i - 1];
            int k = j + (1 << (i - 1));
            if(k <= n)
            {
                if(p[mx[j][i]].w < p[mx[k][i - 1]].w) mx[j][i] = mx[k][i - 1];
                if(p[mi[j][i]].w > p[mi[k][i - 1]].w) mi[j][i] = mi[k][i - 1];
            }
        }
}
int rmqmax(int l, int r)
{
    int m = (int)(log((r - l + 1) * 1.0) / log(2.0));
    if(p[mx[l][m]].w < p[mx[r - (1 << m) + 1][m]].w) return mx[r - (1 << m) + 1][m];
    else return mx[l][m];
}
void solve(int s, int t)
{
    if(s > t) return;
    int pos = rmqmax(s, t);
    printf("(");
    solve(s, pos - 1);
    printf("%s/%d", p[pos].s, p[pos].w);
    solve(pos + 1, t);
    printf(")");
}
int main()
{
    char s[111], tmp[111];
    while(scanf("%d", &n) != EOF && n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%s", s);
            int j;
            for(j = 0; s[j]; j++)
                if(s[j] == '/') break;
                    else tmp[j] = s[j];
            tmp[j++] = '\0';
            int num = 0;
            while(s[j]) {num = num * 10 + s[j] - '0'; j++;}
            strcpy(p[i].s, tmp);
            p[i].w = num;
        }
        sort(p + 1, p + n + 1);
        rmqinit();
        solve(1, n);
        printf("\n");
    }
    return 0;
}

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