ZOJ 3469 Food Delivery [区间DP]

题目连接

题解

简单DP。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define maxn 1005#define smin(x,y) (x)=min((x),(y))using namespace std;int dp[maxn][maxn][2],sum[maxn],n,id,v,pos,tf,ts;struct data{    int x,w;    bool operator < (const data &rhs)const{        return x<rhs.x;    }}my[maxn];int main(){    while(cin>>n>>v>>id){        for(int i=1;i<=n;i++)            cin>>my[i].x>>my[i].w;        my[++n].x=id,my[n].w=0,sum[0]=0;        sort(my+1,my+1+n);        for(int i=1;i<=n;i++){            if(my[i].x==id&&my[i].w==0)pos=i;            sum[i]=sum[i-1]+my[i].w;        }        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                dp[i][j][0]=dp[i][j][1]=1e9;        dp[pos][pos][1]=dp[pos][pos][0]=0;        for(int i=pos;i>=1;i--)            for(int j=pos;j<=n;j++){                if(i==j)continue;                tf=sum[i]+sum[n]-sum[j];                ts=sum[i-1]+sum[n]-sum[j-1];                smin(dp[i][j][0],min(dp[i+1][j][0]+tf*(my[i+1].x-my[i].x),dp[i+1][j][1]+tf*(my[j].x-my[i].x)));                smin(dp[i][j][1],min(dp[i][j-1][1]+ts*(my[j].x-my[j-1].x),dp[i][j-1][0]+ts*(my[j].x-my[i].x)));            }        cout<<min(dp[1][n][0],dp[1][n][1])*v<<endl;    }    return 0;}