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  • 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19 3
    8 3
    4 2
    2 1
    【样例2100 6
    1 1
    2 3
    3 9
    4 27
    5 81
    6 243
    【样例39999 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14
    【样例2100
    【样例3139815
    

    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
    ",dp[h]);
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/PushyTao/p/13144171.html
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