UVA-1632 Alibaba (interval dp+ scrolling array)

Title: There are n pieces of jewelry in a straight line, the location of each piece of jewellery is known, and the first piece of jewelry disappears at ti moment, q can I collect all the jewels? If you can, find the shortest time. The collection can be done instantaneously.

Topic Analysis: Interval DP. DP (i,j,0) indicates the minimum time required to collect the interval (i,j) and stay at the left, and DP (i,j,1) represents the minimum time required to collect the interval (i,j) and stay on the right side. The state transition equation is

DP (i,j,0) =min (DP (i+1,j,0) +t (i+1)-T (i), DP (I+1,J1,) +t (j)-T (i)), DP (i,j,1) =min (DP (i,j-1,0) +t (j)-T (i), DP (i,j-1,1) +t (j)-T (j-1)).

The data size of the problem is large, you can use the scrolling array to optimize the spatial complexity.

The code is as follows:

# include<iostream># include<cstdio># include<cstring># include<algorithm>using namespace        Std;int dp[2][10005][2];int x[10005],t[10005],n;const int inf=0x7fffffff;int main () {while (~SCANF ("%d", &n)) {            for (int i=0;i<n;++i) {scanf ("%d%d", x+i,t+i);        dp[1][i][0]=dp[1][i][1]=dp[0][i][0]=dp[0][i][1]= (t[i]>0) 0:inf; } for (int i=n-2;i>=0;--i) {for (int j=i+1;j<n;++j) {Dp[i&1][j][0]=dp[i&1][j]                [1]=inf; if (dp[(i&1) ^1][j][0]!=inf&&dp[(i&1) ^1][j][0]+x[i+1]-x[i]<t[i]) dp[i&1][j][0]=min                (dp[i&1][j][0],dp[(i&1) ^1][j][0]+x[i+1]-x[i]); if (dp[(i&1) ^1][j][1]!=inf&&dp[(i&1) ^1][j][1]+x[j]-x[i]<t[i]) Dp[i&1][j][0]=min (d                p[i&1][j][0],dp[(i&1) ^1][j][1]+x[j]-x[i]);              if (Dp[i&1][j-1][0]!=inf&&dp[i&1][j-1][0]+x[j]-x[i]<t[j])      Dp[i&1][j][1]=min (Dp[i&1][j][1],dp[i&1][j-1][0]+x[j]-x[i]); if (Dp[i&1][j-1][1]!=inf&&dp[i&1][j-1][1]+x[j]-x[j-1]<t[j]) dp[i&1][j][1]=min (dp[            I&1][j][1],dp[i&1][j-1][1]+x[j]-x[j-1]);        }} if (Dp[0][n-1][1]==inf&&dp[0][n-1][0]==inf) printf ("No solution\n");    else printf ("%d\n", Min (dp[0][n-1][0],dp[0][n-1][1])); } return 0;}

　　

　　

The following is a non-scrolling array:

# include<iostream># include<cstdio># include<cstring># include<algorithm>using namespace Std;int dp[10005][10005][2];int x[10005],t[10005],n;const int inf=1000000000;int main () {    while (~scanf ("%d", &n))    {        for (int i=0;i<n;++i)            scanf ("%d%d", x+i,t+i), dp[i][i][0]=dp[i][i][1]=0;        for (int i=n-2;i>=0;--i) {for            (int j=i+1;j<n;++j) {                dp[i][j][0]=min (dp[i+1][j][0]+x[i+1]-x[i],dp[i+1] [j] [1]+x[j]-x[i]);                if (Dp[i][j][0]>=t[i]) Dp[i][j][0]=inf;                Dp[i][j][1]=min (Dp[i][j-1][0]+x[j]-x[i],dp[i][j-1][1]+x[j]-x[j-1]);                if (Dp[i][j][1]>=t[j]) dp[i][j][1]=inf;            }        }        if (dp[0][n-1][1]==inf&&dp[0][n-1][0]==inf) printf ("No solution\n");        else printf ("%d\n", Min (dp[0][n-1][0],dp[0][n-1][1]));    }    return 0;}

　　

UVA-1632 Alibaba (interval dp+ scrolling array)