UVA 10564-paths through the Hourglass (DP)

The main idea: to give an hourglass, including an inverted triangle and a triangle, each square has a number 0 to 9, from the top of a lattice to go to the bottom of a lattice, to obtain a special value of the path of how many, and output the smallest dictionary order. The path is represented by a starting point and a series of characters ' L ' and ' R '.

Use A[i][j] to denote the number in column J of line I. Because the dictionary order is required to be the smallest, recursion is pushed from the bottom up. The D[i][j][u] indicates how many strips (i,j) and Paths to u, which are inverted triangles in the upper part, can be summed up by d[i+1][j][u-a[i][j] and D[i+1][j+1][u-a[i][j], and the lower half is a positive triangle, which can be d[i +1][J-1][U-A[I][J]] and D[i+1][j][u-a[i][j]] are added recursively.

Finally, find the most left-most starting point and push back the last line to get a path with the smallest dictionary order.

State transition equation:

D[I][J]=D[I+1][J][U-A[I][J]]+D[I+1][J+1][U-A[I][J]] (upper part)

D[I][J]=D[I+1][J-1][U-A[I][J]]+D[I+1][J][U-A[I][J]] (lower half)

#include <stdio.h> #include <stdlib.h>typedef long long ll;int a[45][45]; LL D[45][45][510];char ans[45];int Main (void) {int i,j,u,v,p,n,m,sit; LL sum;scanf ("%d%d", &n,&m); while ((n!=0) | | (m!=0)) {for (i=1;i<=n;i++) {for (j=1;j<=n-i+1;j++) {scanf ("%d", &a[i][j]);}} for (i=2;i<=n;i++) {for (j=1;j<=i;j++) {scanf ("%d", &a[i+n-1][j]);}} p=9* (2*n-1); if (m>p) {printf ("0\n\n");} Else{for (j=1;j<=n;j++) {d[2*n-1][j][a[2*n-1][j]]=1;} for (i=2*n-2;i>=n;i--) {for (j=1;j<=i-n+1;j++) {for (u=a[i][j];u<=p;u++) {d[i][j][u]=d[i+1][j][u-a[i][j]]+d [I+1] [J+1] [U-a[i][j];}}} for (i=n-1;i>=1;i--) {for (u=a[i][1];u<=p;u++) {d[i][1][u]=d[i+1][1][u-a[i][1]];} for (u=a[i][n-i+1];u<=p;u++) {d[i][n-i+1][u]=d[i+1][n-i][u-a[i][n-i+1]];} for (j=2;j<n-i+1;j++) {for (u=a[i][j];u<=p;u++) {d[i][j][u]=d[i+1][j-1][u-a[i][j]]+d[i+1][j][u-a[i][j]];}}} Sum=0;for (j=1;j<=n;j++) {sum=sum+d[1][j][m];} if (sum==0) {printf ("0\n\n");} else{printf ("%lld\n", sum); for (j=1;j<=n;j++) {if (d[1][j][m]!=0) {Sit=v=j;breaK;}} for (i=1;i<=n-1;i++) {m=m-a[i][v];if (v==1) {ans[i]= ' R ';} else if (v==n-i+1) {ans[i]= ' L '; v=v-1;} Else{if (d[i+1][v-1][m]!=0) {v=v-1;ans[i]= ' L ';} else{ans[i]= ' R ';}}} for (i=n;i<2*n-1;i++) {m=m-a[i][v];if (d[i+1][v][m]!=0) {ans[i]= ' L ';} else{v=v+1;ans[i]= ' R ';}} ans[2*n-1]=0;printf ("%d%s\n", sit-1,ans+1);} for (i=1;i<=2*n;i++) {for (j=1;j<=2*n;j++) {for (u=0;u<=p;u++) {d[i][j][u]=0;}}}} scanf ("%d%d", &n,&m);} return 0;}

UVA 10564-paths through the Hourglass (DP)