Crested Ibis vs Monster——AT动态规划思想

题目描述

Ibis is fighting with a monster.
The health of the monster is H.
Ibis can cast N kinds of spells. Casting the i-th spell decreases the monster’s health by Ai, at the cost of Bi Magic Points.
The same spell can be cast multiple times. There is no way other than spells to decrease the monster’s health.
Ibis wins when the health of the monster becomes 0 or below.
Find the minimum total Magic Points that have to be consumed before winning.

Constraints
·1≤H≤104
·1≤N≤103
·1≤Ai≤104
·1≤Bi≤104
·All values in input are integers.
输入
Input is given from Standard Input in the following format:

H N
A1 B1
:
AN BN

输出

Print the minimum total Magic Points that have to be consumed before winning.

样例输入

【样例1】
9 3
8 3
4 2
2 1
【样例2】
100 6
1 1
2 3
3 9
4 27
5 81
6 243
【样例3】
9999 10
540 7550
691 9680
700 9790
510 7150
415 5818
551 7712
587 8227
619 8671
588 8228
176 2461

样例输出

【样例1】
4
【样例2】
100
【样例3】
139815

提示

样例1解释
First, let us cast the first spell to decrease the monster’s health by 8, at the cost of 3 Magic Points. The monster’s health is now 1.
Then, cast the third spell to decrease the monster’s health by 2, at the cost of 1 Magic Point. The monster’s health is now −1.
In this way, we can win at the total cost of 4 Magic Points.
样例2解释
It is optimal to cast the first spell 100 times.

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
///#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
///char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
int num[maxn];
ll ans;
int main() {
    int h=read,n=read;
    vector<int> dp(h+1, mod);
    dp[0]=0;
    for(int i=0;i<n;i++)
    {
        int a=read,b=read;
        for(int j=0;j<h;j++)
        {
            int temp=min(j+a,h);
            dp[temp] = min(dp[temp],dp[j]+b);
        }
    }
    printf("%d\n",dp[h]);
    return 0;
}