[usaco2004][bzoj3379] 交作业

按距离从小到大排序

f[i][j][0或1]表示在i或j还有i-j没有完成

转移

    tmp=dp[i][j][0];
    tmp=min(tmp,max(dp[i][j+1][1]+a[j+1].dist-a[i].dist,a[i].t));
    tmp=min(tmp,max(dp[i-1][j][0]+a[i].dist-a[i-1].dist,a[i].t));

注意边界 比如:dp[0][i],dp[0][c+3]初值应为inf

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+30,inf=1e6;
int dp[N][N][2];
#define ri register int 
int az1,az2,tmp;
int c,b,h;
struct re{
    int dist;
    int t;
};
re a[N];
bool cmp(re x,re y){return x.dist<y.dist;}
//int max(int x,int y){if(x>y)return x;else return y;}
//int min(int x,int y){if(x<y)return x;else return y;}
int main()
{    //freopen("p.in","r",stdin);
    cin>>c>>h>>b;
    for(int i=1;i<=c;i++)
    cin>>a[i].dist>>a[i].t;
    sort(a+1,a+1+c,cmp);
    for(int i=0;i<N;i++)
     for(int j=0;j<N;j++)dp[i][j][0]=dp[i][j][1]=inf;
     int sdr5s;
    dp[1][c][0]=max(a[1].t,a[1].dist);
    dp[1][c][1]=max(a[c].t,a[c].dist);
    for(int i=1;i<=c;i++)
      for(int j=c;j>=i;j--){
          tmp=dp[i][j][0];
          tmp=min(tmp,max(dp[i][j+1][1]+a[j+1].dist-a[i].dist,a[i].t));
          tmp=min(tmp,max(dp[i-1][j][0]+a[i].dist-a[i-1].dist,a[i].t));
        dp[i][j][0]=tmp;
        tmp=dp[i][j][1];
        tmp=min(tmp,max(dp[i][j+1][1]+a[j+1].dist-a[j].dist,a[j].t));
          tmp=min(tmp,max(dp[i-1][j][0]+a[j].dist-a[i-1].dist,a[j].t));
          dp[i][j][1]=tmp;
      }
      az1=1e8;
    for(int i=1;i<=c;i++)az1=min(az1,(min(dp[i][i][0],dp[i][i][1])+(int)fabs(b-a[i].dist)));
    cout<<az1;
    return 0;
}
View Code

 

posted @ 2018-08-30 20:38  周栎  阅读(...)  评论(...编辑  收藏