当时看题以为需要好多种分类讨论,看到题解的dp简单机智...(还是自己dp没学会)
dp用的妙
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 5e3 + 10;
const int INF = 1e9;
ll n, m, k, a, b, c;
int T;
ll f[maxn];
void solve()
{
memset(f, 0, sizeof(f));
f[0] = 0;
for(int i = 1; i <= max(n + k, m + k); i++)
{
f[i] = (ll)maxn * INF;
if(i >= 2) f[i] = min(f[i], min(f[i - 2] + a, f[i - 2] + b));
if(i >= 3) f[i] = min(f[i], f[i - 3] + b);
if(i >= 1) f[i] = min(f[i], f[i - 1] + min(a, b));
}
}
int main()
{
cin >> T;
while(T--)
{
ll ans = (ll)maxn * INF; //注意初始化问题,hdu1024也WA在了初始化
cin >> n >> m >> k >> a >> b >> c;
solve(); //计算同性别各种人数花费最小
for(int i = 0; i <= k; i++)
{
ll t = i * c + f[n + k - i] + f[m + k - i]; //按情侣对数情况遍历讨论,简单好理解
ans = min(ans, t);
}
cout << ans << endl;
}
return 0;
}