GarsiaWachs算法

对于一列的石子归并问题，除了朴素的O(n^3)的dp做法及其O(n^2)优化，还有GarsiaWachs算法。

算法流程是，找一个最小的k，使得a[k-1]<=a[k+1]，将a[k-1]和a[k]合并；从当前位置向前找到一个最大的i，使得a[i]>a[k-1]+a[k]，并将新合并的一堆移到i的后面；重复操作n-1次，直至只剩下1堆，答案就是每次合并结果累加起来。

#include <cstdio>

const int maxn = 105, inf = 0x3f3f3f3f;

struct Node {
    int num, pre, next;
} a[maxn];

inline void del(int x) {
    int pp = a[a[x].pre].pre, nn = a[x].next;
    a[pp].next = nn, a[nn].pre = pp;
}

inline void insert(int x, int y) {
    a[x].pre = y, a[x].next = a[y].next;
    a[a[y].next].pre = x, a[y].next = x;
}

int main() {
    int n, ans = 0;
    scanf("%d", &n);
    a[0].num = a[n + 1].num = inf;
    a[0].next = 1, a[n + 1].pre = n;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i].num);
        a[i].pre = i - 1, a[i].next = i + 1;
    }
    for (int t = 1; t <= n - 1; ++t) {
        int x, y, sum;
        for (int i = a[0].next; ; i = a[i].next)
            if (a[a[i].pre].num <= a[a[i].next].num) {
                x = i;
                break;
            }
        sum = a[a[x].pre].num + a[x].num;
        for (int i = a[a[x].pre].pre; ; i = a[i].pre)
            if (a[i].num > sum) {
                y = i;
                break;
            }
        del(x);
        a[x].num = sum;
        printf("%d ttt\n", sum);
        insert(x, y);
        ans += sum;
    }
    printf("%d", ans);
    return 0;
}