HDU1160FatMouse's Speed

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#include<stack>
using namespace std;
const int maxn=1010;
struct point
{
    int w,s,per,id;
}p[maxn];
int cmp(point a,point b)
{
    return a.s>b.s;
}
int dp[maxn];
int main()
{
    int i,j,k,n,m;
    n=1;
    memset(dp,0,sizeof(dp));
    while(scanf("%d%d",&p[n].w,&p[n].s)!=EOF)
    {
        p[n].id=n;
        p[n].per=0;
        n++;
     }
     sort(p+1,p+n,cmp);
     int max;
     dp[1]=1;//dp[i]记录以第i个元素结尾的最长递增的长度
     for(i=2;i<n;i++)
     {
         max=0;
         for(j=1;j<i;j++)
         {
             if(p[j].w<p[i].w && p[j].s>p[i].s)
             {
                 if(dp[j]>=max)
                 {
                     max=dp[j];
                     p[i].per=j;//记录i位置的前驱
                 }
             }
             dp[i]=max+1;
         }
     }
     max=-0x3f3f3f3f;
     int end;
    for(i=1;i<n;i++)
    {
        if(max<=dp[i])
        {
            max=dp[i];
            end=i;
        }
    }
     printf("%d
",max);
     stack<int> ok;
     while(end!=0)
     {
         ok.push(end);
         end=p[end].per;
     }
     while(!ok.empty())
     {
         printf("%d
",p[ok.top()].id);
         ok.pop();
     }
     return 0;
}


/*二分写法*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
struct X{
    int s,w,sb;
};
bool cmp(X a,X b)
{
    return (a.s>b.s)||(a.s==b.s&&a.w>b.w);//这里排序注意了，由于本题数据水的不能再水，a.w<b.w也过了，正解应该是a.w>b.w
}
X xx[1005];
int pre[1005];
int fnm[1005];
int shuchu[1005];
int main()
{
    int i=0,n=1,w,s;
    while(scanf("%d%d",&w,&s)!=EOF)
    {
        xx[n].w=w;
        xx[n].s=s;
        xx[n].sb=n++;
    }
    n--;
    sort(xx+1,xx+n+1,cmp);/*
    for(i=1;i<=n;i++)
        printf("%d %d
",xx[i].w,xx[i].s);*/
    int lsb=0,j,l,r;
    fnm[0]=1;
    pre[fnm[0]]=-1;
    for(i=2;i<=n;i++)
    {
        l=0,r=lsb;
        while (l<=r) //二分查找
        {
            int mid = (l+r)>>1;
            if (xx[fnm[mid]].w < xx[i].w)
                l=mid+1;
            else
                r=mid-1;
        }
        j=l;
        if(j>lsb)
            lsb++;
        fnm[j]=i;//存下标，相当于c[j]=a[i];
        if(j!=0)//fnm小标从0开始，所以要判断第一个
            pre[i]=fnm[j-1];//j-1的元素肯定是最长递增中的元素
        else
            pre[i]=-1;
    }
    printf("%d
",lsb+1);
    int xsb=fnm[lsb];
    i=0;
    while(pre[xsb]!=-1)
    {
        shuchu[i++]=xx[xsb].sb;
        xsb=pre[xsb];
    }
    printf("%d
",xx[xsb].sb);
    for(i=lsb-1;i>=0;i--)
        printf("%d
",shuchu[i]);
    return 0;
}