将第一次正式的转载献给队友酱,代码部分还是有些失败
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,注意数据范围,仅开成INF是不行的
- #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;
- }