Hdu 4960 Another OCD Patient (memory)

Hdu 4960 Another OCD Patient (memory)

Link: hdu 4960 Another OCD Patient

Given a sequence with a length of n, and then giving n numbers of ai, it indicates the cost of merging I numbers. Each time, a continuous subsequence can be converted into a number, that is, the sum of each item in the sequence. It is required that a sequence with a given length of n be converted into a return string, and a number can only be merged once.

Solution: dp [l] [r] indicates the minimum cost from l to r to the input string. dp [l] [r] = min (val (r? L + 1), val (r? I + 1) + val (j? L + 1) + dp [j + 1] [I? 1]), when I decreases by 1, the corresponding j must increase, which can reduce the majority of enumeration.

#include 
  
   #include 
   
    #include using namespace std;typedef long long ll;const int maxn = 5005;int N, arr[maxn], dp[maxn][maxn];ll x, val[maxn];inline ll getval (int l, int r) {    return val[r] - val[l-1];}void init () {    memset(dp, -1, sizeof(dp));    val[0] = 0;    for (int i = 1; i <= N; i++) {        scanf("%I64d", &x);        val[i] = val[i-1] + x;    }    for (int i = 1; i <= N; i++)        scanf("%d", &arr[i]);}int solve (int l, int r) {    if (l > r)        return 0;    if (dp[l][r] != -1)        return dp[l][r];    int& ret = dp[l][r];    ret = arr[r-l+1];    int mv = l;    for (int i = r; i > l; i--) {        ll u = getval(i, r);        while (getval(l, mv) < u && mv < i)            mv++;        if (mv >= i)            break;        if (getval(l, mv) == u)            ret = min(ret, arr[r-i+1] + arr[mv-l+1] + solve(mv+1, i-1));    }    return ret;}int main () {    while (scanf("%d", &N) == 1 && N) {        init();        printf("%d\n", solve(1, N));    }    return 0;}