POJ 1785 treap 或 RMQ - c++编程基础

[cpp]
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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
#include
#include
#include
#include
#include
#include
#include
#include
#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;
}

然后还可以使用RMQ来搞

还是首先要排序

每次递归处理一个区间，找出有最大第二关键字的那个做根。他左边的就是他的左子树，右边的是右子树

POJ 1785 treap 或 RMQ (一)