2012暑假集训内部测试赛3

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2427

线段树+离散化  不离散化不知道会不会超时 一直RE 可能N值没有说的那么小吧 题意有问题 按1W开数组就RE 按10W开就A了

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 100001
int s[N*6],num,f[100011],po[100011][2];
struct node
{
    int li,num;
}line[200011];
void build(int l,int r,int w)
{
    s[w] = -1;
    if(l==r)
    {
        return ;
    }
    int m = (l+r)/2;
    build(l,m,2*w);
    build(m+1,r,2*w+1);
}
void add(int a,int b,int da,int l,int r,int w)
{
    if(l==a&&b==r)
    {
        s[w] = da;
        return ;
    }
    int m = (l+r)/2;
    if(s[w]>0)
    {
        s[2*w] = s[w];
        s[2*w+1] = s[w];
        s[w] = -1;
    }
    if(b<=m)
        add(a,b,da,l,m,2*w);
    else
        if(a>m)
            add(a,b,da,m+1,r,2*w+1);
        else
        {
            add(a,m,da,l,m,2*w);
            add(m+1,b,da,m+1,r,2*w+1);
        }
}
void search(int l,int r,int w)
{
    if(s[w]>0)
    {
        f[s[w]] = 1;
        return ;
    }
    if(l==r)
        return ;
    int m = (l+r)/2;
    search(l,m,2*w);
    search(m+1,r,2*w+1);    
}
bool cmp(node a,node b)
{
    return a.num<b.num;
}
int main()
{
    int n,i,j,t,a,b;
    scanf("%d", &t);
    while(t--)
    {
        memset(f,0,sizeof(f));    
        num = 0;        
        scanf("%d", &n);
        build(1,N,1);    
        for(i = 0; i < n ;i++)
        {
            scanf("%d%d", &po[i][0],&po[i][1]);
            line[2*i].li = i+1;
            line[2*i].num = po[i][0];
            line[2*i+1].li = -(i+1);
            line[2*i+1].num = po[i][1];
        }
        sort(line,line+2*n,cmp);    
        int te = line[0].num,g = 1;
        for(i = 0 ; i < 2*n ; i++)
        {
            if(te!=line[i].num)
            {
                te = line[i].num;
                g++;
            }
            if(line[i].li>0)
            {
                po[line[i].li][0]=g;
            }
            else
                po[-line[i].li][1]=g;
        }
        for(i = 1; i <= n ; i++)
        {
            add(po[i][0],po[i][1],i,1,N,1);
        }
        search(1,N,1);
        for(i = 1; i <= n ; i++)
            if(f[i])

                num++;
        printf("%d\n",num);
    }
    return 0;
} 

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2430

DP 前一个的1或者最高 到这一个1或者最高 中间选一个最优的

#include<iostream>
#include<cstdio>
#include<string.h>
#include<cmath>
using namespace std;
int main()
{
    int i,j,n,m,x,a[111],k;
    double dp[111][111];
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= n ; i++)
        scanf("%d",&a[i]);
        scanf("%d", &k);
        if(n==1)
        {
            printf("0.000000\n");
            continue;
        }
        for(i = 2; i <= n ; i++)
        {
            x = a[i-1];
            if(dp[i][1]<dp[i-1][x]+sqrt(k*k+(x-1)*(x-1)))
            dp[i][1] = dp[i-1][x]+sqrt(k*k+(x-1)*(x-1));
            if(dp[i][1]<dp[i-1][1]+k)
            dp[i][1] = dp[i-1][1]+k;
            if(dp[i][a[i]]<dp[i-1][x]+sqrt(k*k+(x-a[i])*(x-a[i])))
            dp[i][a[i]] = dp[i-1][x]+sqrt(k*k+(x-a[i])*(x-a[i]));
            if(dp[i][a[i]]<dp[i-1][1]+sqrt(k*k+(a[i]-1)*(a[i]-1)))
            dp[i][a[i]] = dp[i-1][1]+sqrt(k*k+(a[i]-1)*(a[i]-1));
        }
        if(dp[n][1]>dp[n][a[n]])
        printf("%.6lf\n",dp[n][1]);
        else
        printf("%.6lf\n",dp[n][a[n]]);
    }
    return 0;
}
 

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2429

模拟 这题WA惨了 最后两分钟交对 好险。。

x不能为负值 就算最后有符合的负值也是输出-1 考虑两种情况x为0或者不为0的情况

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,n,f[401],m;
    long long  k,a[61],stack[401],b[61];
    while(scanf("%d",&n)!=EOF)
    {
        int top =200;
        memset(stack,0,sizeof(stack));
        memset(f,0,sizeof(f));
        for(i = 1; i <= n ; i++)
        {
            scanf("%lld",&a[i]);
            b[i] = a[i];
        }
        scanf("%lld", &k);
        for(j = 1; j <= n ; j++)
        {
            if(a[j]==-1)
                a[j]=0;
            if(a[j]==0)
            {
                top--;
                long long  x = stack[top];
                top--;
                x+=stack[top];
                stack[top++] = x;
            }
            else
                stack[top++] = a[j];
        }
        top--;
        if(stack[top]==k)
        {
            printf("0\n");
            continue;
        }
        top = 200;
        for(j = 1; j <= n ; j++)
        {
            if(b[j]==-1)
            {
                f[top]=1;
                stack[top++] = 0;
            }
            else
            if(b[j]==0)
            {
                top--;
                long long  x = stack[top];
                top--;
                x+=stack[top];
                if(f[top]||f[top+1])
                {
                    f[top] = 1;
                }
                stack[top++] = x;        
            }
            else
                stack[top++] = b[j];
        }
        top--;
        if(!f[top]&&stack[top]==k)
        {
            printf("0\n");
            continue;
        }
        if(f[top]&&k-stack[top]>0)
            printf("%lld\n",k-stack[top]);
        else
            printf("-1\n");
    }
    return 0;
}