Fzu 1005 Fast Food (DP)

Test instructions: Input n,k, and then enter n number, this n number represents the distance from the starting point, now to select K points, so that the distance and the smallest.

Analysis: We ask to choose K points from n points to minimize the distance, that is, dp[n][k], then dp[n][k] is Dp[n-1][k-1]+sum[n][n],dp[n-2][k-1]+sum[n-1][n],dp[n-3][k-1]+sum [N-2] The minimum value between [n]....dp[1][k-1]+sum[2][n].

where sum[x][y] means to select a point from the X-point to the Y-point at the shortest distance (if the number is singular, select the middle point, even if the middle of the choice of two whichever line)

Memory Search:

#include <iostream>#include<queue>#include<cstdio>#include<cstring>#include<cstdlib>#include<stack>using namespacestd;Const intOO =1e9;inta[222];intdp[222][ -];intsum[222][222];intDFS (intNintk) {    if(n<0|| k<0)        returnOo; if(dp[n][k]!=-1)        returnDp[n][k]; if(k>=N) {dp[n][k]=0; return 0; } Dp[n][k]=Oo;  for(intI=1; i<=n; i++) {Dp[n][k]= Min (Dp[n][k], DFS (i-1, K-1)+sum[i][n]);//For the first n points to establish a K-point, can be from the first 1 to n-1 points to establish k-1 points pushed to find the best solution. }    returndp[n][k];}intMain () {intN, M; intKK =1;  while(SCANF ("%d%d", &n, &m), n+m) {a[0] =0;  for(intI=1; i<=n; i++) {scanf ("%d", &A[i]); }         for(intI=1; i<=n; i++) {Sum[i][i]=0;  for(intj=i+1; j<=n; J + +) {Sum[i][j]= sum[i][j-1]+a[j]-a[(I+J)/2];//sum[i][j] means to select a point between I,j, the shortest distance consumed (if the singular, select the middle point, even if the middle of the selection of two)}}} memset (DP,-1,sizeof(DP)); intAns =DFS (n, m); printf ("Chain%d\n", kk++); printf ("Total Distance sum =%d\n\n", ans); }    return 0;}

Dp

#include <iostream>#include<queue>#include<cstdio>#include<cstring>#include<cstdlib>#include<stack>using namespacestd;Const intOO =1e9;inta[222];intdp[222][ -];intsum[222][222];intMain () {intN, M; intKK =1;  while(SCANF ("%d%d", &n, &m), n+m) {a[0] =0;  for(intI=1; i<=n; i++) {scanf ("%d", &A[i]); }         for(intI=1; i<=n; i++) {Sum[i][i]=0;  for(intj=i+1; j<=n; J + +) {Sum[i][j]= sum[i][j-1]+a[j]-a[(I+J)/2]; }        }         for(intI=0; i<=n; i++)        {             for(intj=0; j<=m; J + +) {Dp[i][j]=Oo; }} dp[0][0] =0;  for(intI=1; i<=n; i++)        {             for(intk=1; k<=m; k++)            {                 for(intj=0; j<i; J + +)//This layer loops and upper loop which can be outside, because Dp[i][k] only and dp[1][k-1] to dp[i-1][k-1] about {dp[i][k]= Min (Dp[i][k], dp[j][k-1]+sum[j+1][i]);//Enumerate the state from 1 to i-1, update the value of Dp[i][k] }}} printf ("Chain%d\n", kk++); printf ("Total Distance sum =%d\n\n", Dp[n][m]); }    return 0;}

Fzu 1005 Fast Food (DP)