Stone merging problem (naïve interval dp&&garsiawachs algorithm)

Topic Links:

http://poj.org/problem?id=1738

Given n heap of stones, each time can only merge adjacent two piles of stone, each time the charge is the two piles of stone and;

Method One:

Interval DP complexity of O (n^3)

State transition equation

Dp[i][j]=min (Dp[i][k]+dp[k+1][j]+sup[i][j])

DP[I][J] represents the minimum cost of merging from I heap stones to the J heap of Stones, sum[i][j] represents from the I heap to the J Heap of Stones and

The code is as follows:

<span style= "FONT-SIZE:14PX;" > #include <iostream> #include <cstring> #include <cstdio>using namespace Std;const int maxn = 110; typedef long Long LL; LL SUM[MAXN]; LL DP[MAXN][MAXN];   LL A[maxn];int Main () {int n;        while (~SCANF ("%d", &n)) {memset (sum,0,sizeof (sum));            for (int i=0;i<n;i++) {scanf ("%lld", &a[i]);            if (i==0) sum[i]=a[i];        else Sum[i]=sum[i-1]+a[i];        } for (int i=0;i<maxn;i++) for (int j=i+1;j<maxn;j++) dp[i][j]=1000000000000000000;                for (int len=2;len<=n;len++) {for (int i=0;i+len<=n;i++) {int j = i+len-1; for (int k=i;k<j;k++) {if (i==0) {dp[i][j]=min (Dp[i][k]+dp[k+1][j]+sum[j]                        , Dp[i][j]);                        printf ("dp[%d][%d]%d", i,k,dp[i][k]);                        printf ("dp[%d][%d]%d\n", k+1,j,dp[k+1][j]); cout<< "ans" <<dp[i][j]<<endl;                } else Dp[i][j]=min (Dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1],dp[i][j]);   }}} cout<<dp[0][n-1]<<endl; } return 0;} </span>

Method Two:

Garsiawachs algorithm

1) Set the stone sequence to st[], from left to right to find a minimum k to make a[k-1]<=a[k+1], merging K and K-1 Heap

2) then go left and find one of the biggest J make A[j]>a[k]+a[k-1] and put A[k-1]+a[k in his back

3) Repeat this operation until there is only a pile left.

Note: Set A[-1],a[n] as positive infinity

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring>using namespace std;const int maxn = 50010;int    Stone[maxn],n,t;int ret; void combine (int k) {int tmp = Stone[k] + stone[k-1];    RET +=tmp;    for (int i=k; i<t-1; i++) {stone[i] = stone[i+1];    } t--;    Int J;    for (j=k-1; j>0 && stone[j-1]<tmp; j--) {stone[j] = stone[j-1];    } Stone[j] = tmp;        while (j>=2 && stone[j]>=stone[j-2]) {int d = t-j;        Combine (j-1);    j = t-d;        }}int Main () {while (scanf ("%d", &n), N) {for (int i=0; i<n; i++) scanf ("%d", stone+i);        t=1,ret=0;            for (int i=1; i<n; i++) {stone[t++] = Stone[i];            while (t>=3 && stone[t-3]<=stone[t-1]) {combine (t-2);        }} while (t>1) combine (t-1);    printf ("%d\n", ret); } return 0;}

Stone merging problem (naïve interval dp&&garsiawachs algorithm)