HDU 1025 Constructing Roads In JGShining's Kingdom（LIS的O(nlogn)算法）
题目地址：点击打开链接
题意：一条河的2边住着穷人和富人，一个富人只打算和一个特定穷人之间建一座桥，问最多能建多少个桥
思路：O(n^n)的会超时，主要思想看后面大神解析
错误代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
using namespace std;
const int maxn = 500010;
struct node
{
int poor,rich;
}a[maxn];
int dp[maxn];
bool cmp(node x,node y)
{
return x.poor < y.poor;
}
int main()
{
int n,i,j,cas = 1;
while(scanf("%d",&n) != EOF)
{
for(i=0; i<n; i++)
{
scanf("%d%d",&a[i].poor,&a[i].rich);
dp[i] = 1;
}
sort(a,a+n,cmp);
for(i=1; i<n; i++)
{
int max1 = 0,p = 0;
for(j=0; j<i; j++)
{
if(a[i].rich > a[j].rich && dp[j] > max1)//这句话每次都会执行，导致dp[i]是错的
{
max1 = dp[j];
p = j;
}
}
if(p != 0)//i==1的时候p正好为0
{
dp[i] = max1 + 1;
}
}
int max2 = 0;
for(i=0; i<n; i++)
{
if(dp[i] > max2)
max2 = dp[i];
}
printf("Case %d:\n",cas++);
if(max2 == 1)
{
printf("My king, at most 1 road can be built.\n\n");
}
else
{
printf("My king, at most %d roads can be built.\n\n",max2);
}
}
return 0;
}
逻辑比较混乱，写错了，重写了一个
超时代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
using namespace std;
const int maxn = 500010;
struct node
{
int poor,rich;
}a[maxn];
int dp[maxn];
bool cmp(node x,node y)
{
return x.poor < y.poor;
}
int main()
{
int n,i,j,cas = 1;
while(scanf("%d",&n) != EOF)
{
memset(dp,0,sizeof(dp));
for(i=0; i<n; i++)
{
scanf("%d%d",&a[i].poor,&a[i].rich);
}
sort(a,a+n,cmp);
dp[0] = 1;
for(i=1; i<n; i++)
{
int max1 = 0;
for(j=0; j<i; j++)
{
if(a[i].rich > a[j].rich && dp[j] > max1)
{
max1 = dp[j];
}
}
dp[i] = max1 + 1;//没有符合条件的就为1，有符合条件的就加1
}
int max2 = 0;
for(i=0; i<n; i++)
{
if(dp[i] > max2)
max2 = dp[i];
}
printf("Case %d:\n",cas++);
if(max2 == 1)
{
printf("My king, at most 1 road can be built.\n\n");
}
else
{
printf("My king, at most %d roads can be built.\n\n",max2);
}
}
return 0;
}
思路正确，超时了
AC代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
using namespace std;
const int maxn = 500010;
int a[maxn],stack1[maxn];
int main()
{
int n,i;
int poor,rich,cas = 1;
while(scanf("%d",&n) != EOF)
{
for(i=0; i<n; i++)
{
scanf("%d%d",&poor,&rich);
a[poor] = rich;
}
int top = 0;
stack1[0] = 0;
for(i=1; i<=n; i++)
{
if(a[i] > stack1[top])
{
stack1[++top] = a[i];
}
else
{
int low = 1,high = top,mid;
while(low <= high)
{
mid = (low + high) / 2;
if(a[i] > stack1[mid])
{
low = mid + 1;
}
else
{
high = mid - 1;
}
}
stack1[low] = a[i];//里面是low,high会导致错误，细节多注意
}
}
printf("Case %d:\n",cas++);
if(top == 1)
{
printf("My king, at most 1 road can be built.\n\n");
}
else
{
printf("My king, at most %d roads can be built.\n\n",top);
}
}
return 0;
}
看了大神的思想A的，
大神地址1：点击打开链接
大神地址2：点击打开链接