poj1738 An old Stone Game 【GarsiaWachs算法】【动态规划】经典合并石子问题

 

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

题目大意：有n堆石子排在一行，每堆石子给定一个重量。要把n堆石子合并成一堆，每次合并智能将相邻的两堆石子合并成一堆，代价为这两堆石子的重量和，求最小的总代价。

参考资料：

http://wenku.baidu.com/view/84c326fb700abb68a982fbcc.html

http://fanhq666.blog.163.com/blog/static/81943426201062865551410/

http://hi.baidu.com/nyroro/item/4ad998fc1234cad242c36a86

下面是GarsiaWachs算法的代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
const int N = 50050;
int n , t , stone[N] , ans;
void combine(int k) {
    int tmp = stone[k] + stone[k-1];
    ans += 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 , ans = 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" , ans);
    }
    return 0;
}