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\n",max);     stack<int> ok;     while(end!=0)     {         ok.push(end);         end=p[end].per;     }     while(!ok.empty())     {         printf("%d\n",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\n",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\n",lsb+1);    int xsb=fnm[lsb];    i=0;    while(pre[xsb]!=-1)    {        shuchu[i++]=xx[xsb].sb;        xsb=pre[xsb];    }    printf("%d\n",xx[xsb].sb);    for(i=lsb-1;i>=0;i--)        printf("%d\n",shuchu[i]);    return 0;  }