POJ

Everybody is fond of computers, but buying a new one is always a money challenge. Fortunately, there is always a convenient way to deal with. You can replace your computer and get a brand new one, thus saving some maintenance cost. Of course, you must pay a fixed cost for each new computer you get.

Suppose you are considering an n year period over which you want to have a computer. Suppose you buy a new computer in year y, 1<=y<=n Then you have to pay a fixed cost c, in the year y, and a maintenance cost m(y,z) each year you own that computer, starting from year y through the year z, z<=n, when you plan to buy - eventually - another computer.

Write a program that computes the minimum cost of having a computer over the n year period.

Input

The program input is from a text file. Each data set in the file stands for a particular set of costs. A data set starts with the cost c for getting a new computer. Follows the number n of years, and the maintenance costs m(y,z), y=1..n, z=y..n. The program prints the minimum cost of having a computer throughout the nyear period.

White spaces can occur freely in the input. The input data are correct and terminate with an end of file.

Output

For each set of data the program prints the result to the standard output from the beginning of a line.

Sample Input

3
3
5 7 50
6 8
10

Sample Output

19

Hint

An input/output sample is shown above. There is a single data set. The cost for getting a new computer is c=3. The time period n is n=3 years, and the maintenance costs are:

• For the first computer, which is certainly bought: m(1,1)=5, m(1,2)=7, m(1,3)=50,
• For the second computer, in the event the current computer is replaced: m(2,2)=6, m(2,3)=8,
• For the third computer, in the event the current computer is replaced: m(3,3)=10.

思路：用dp[i]代表前i天的最小花费，则对于每个j >= i， dp[j] = min(dp[j], dp[i - 1] + c + cost[i][j - i + 1])。所以暴力跑个n^2的dp就行了。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int c, n, cost[5015][5015], dp[5015];
int main() {
    while (~scanf("%d%d", &c, &n)) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n - i + 1; ++j) {
                scanf("%d", &cost[i][j]);
            }
        }
        clr(dp, INF);
        dp[0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                dp[j] = min(dp[j], dp[i - 1] + c + cost[i][j - i + 1]);
            }
        }
        printf("%d
", dp[n]);
    }
    return 0;
}