poj 1991 Turning in Homework dp

这个可以证明必须从两边的任务开始交起，因为中间交的任务可以后面经过的时候再交，所以就变成了一个n*n的dp。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e3+9;
int dp[maxn][maxn][2];
struct D
{
    int x,t;
    bool operator <(const D & xx) const
    {
        return x<xx.x;
    }
}a[maxn];

int ff(int x)
{
    if(x<0) return -x;
    return x;
}

int main()
{
    int n,m,b;
    while(scanf("%d %d %d",&n,&m,&b)!=EOF)
    {
        for(int i=1;i<=n;i++)
        scanf("%d %d",&a[i].x,&a[i].t);
        sort(a+1,a+1+n);
        memset(dp,50,sizeof(dp));
        dp[0][n+1][0]=0;
        for(int k=0;k<n;k++)
        for(int i=0;i<=k;i++)
        {
            int tmp=max(dp[i][n+1-(k-i)][0]+a[i+1].x-a[i].x,a[i+1].t);
            dp[i+1][n+1-(k-i)][0]=min(dp[i+1][n+1-(k-i)][0],tmp);

            tmp=max(dp[i][n+1-(k-i)][0]+a[n+1-(k-i)-1].x-a[i].x,a[n+1-(k-i)-1].t);
            dp[i][n+1-(k-i)-1][1]=min(dp[i][n+1-(k-i)-1][1],tmp);

            tmp=max(dp[i][n+1-(k-i)][1]+a[n+1-(k-i)].x-a[n+1-(k-i)-1].x,a[n+1-(k-i)-1].t);
            dp[i][n+1-(k-i)-1][1]=min(dp[i][n+1-(k-i)-1][1],tmp);

            tmp=max(dp[i][n+1-(k-i)][1]+a[n+1-(k-i)].x-a[i+1].x,a[i+1].t);
            dp[i+1][n+1-(k-i)][0]=min(dp[i+1][n+1-(k-i)][0],tmp);
        }
        int ans=1e10;
        for(int i=0;i<=n;i++)
        {
            ans=min(ans,dp[i][i+1][0]+ff(b-a[i].x));
            ans=min(ans,dp[i][i+1][1]+ff(b-a[i+1].x));
        }
        cout<<ans<<endl;
    }
    return 0;
}