UVA 1632-alibaba (DP)

The main idea: in a straight line there are N (n<=10000) points have treasures, each point coordinates are integers, each treasure in a certain time to disappear after, you can start at any point, moving a unit needs a unit of time, ask at least how much time to take out all the treasures, If you cannot take out all the output no solution. The input is incremented in order of coordinates.

A[i] denotes the location of the first treasure, B[i] indicates the time of disappearance.

First notice that for any interval (i,j), after taking all the treasures must be in either I or J. So the d[i][j][0] means to take all the treasures between the first and the first J, and at the end of I, using d[i][j][1] to represent all the treasures between the first and the first J and the last in J.

The program uses a scrolling array to optimize memory consumption, and recursion is in order of increasing the width of the interval, so it is also scrolling in this way.

State transition equation:

d[i][j][0]=min {D[i+1][j][0]+a[i+1]-a[i],d[i+1][j][1]+a[i+j]-a[i]}

d[i][j][1]=min {D[i][j-1][0]+a[i+j]-a[i],d[i][j-1][1]+a[i+j]-a[i+j-1]}

In the program, you also need to determine whether the recursive process can be reached before disappearing, and will be set to INF.

#include <stdio.h> #include <stdlib.h>int a[10010];int b[10010];int d[2][10010][2];int Main (void) {int i,j, N,cur,min;while (scanf ("%d", &n) ==1) {for (i=1;i<=n;i++) {scanf ("%d%d", &a[i],&b[i]);} Cur=0;for (j=1;j<=n;j++) {d[0][j][0]=d[0][j][1]= (b[j]>0)? 0: (1<<30);} for (i=1;i<n;i++) {cur^=1;for (j=1;j<=n-i;j++) {d[cur][j][0]=d[cur^1][j+1][0]+a[j+1]-a[j]>d[cur^1][j+1][1 ]+A[J+I]-A[J]?D[CUR^1][J+1][1]+A[J+I]-A[J]:d [cur^1][j+1][0]+a[j+1]-a[j];d [cur][j][1]=d[cur^1][j][0]+a[j+i]-a[j ]>D[CUR^1][J][1]+A[J+I]-A[J+I-1]?D[CUR^1][J][1]+A[J+I]-A[J+I-1]:d [cur^1][j][0]+a[j+i]-a[j];d [cur][j][0]=d[ CUR][J][0]<B[J]?D[CUR][J][0]:(1<<30);d [cur][j][1]=d[cur][j][1]<b[j+i]?d[cur][j][1]:(1<<30);}} MIN=D[CUR][1][0]>D[CUR][1][1]?D[CUR][1][1]:d [Cur][1][0];if (min== (1<<30)) {printf ("No solution\n");} else{printf ("%d\n", Min);}} return 0;}

UVA 1632-alibaba (DP)