Hdu 4960 Another OCD Patient (memory-based), hdu4960

Hdu 4960 Another OCD Patient (memory-based), hdu4960

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 of being input and input strings from l to r, 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 value increases, which can reduce most of the enumeration values.

#include <cstdio>#include <cstring>#include <algorithm>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;}