Inviting Friends (HDU"/>
Inviting Friends (HDU
一.题目链接:
HDU-3244
二.题目大意:
有 n 种原料,每种原料有 6 个属性.
x:每个人需要此原料的量
y:已有的量
s1:小包原料的数量
p1:小包原料的价钱
s2:大包原料的数量
p2:大包原料的价钱
现有 m 元,求最多的人数.
三.分析:
二分人数,check(mid) 函数里面对每种原料完全背包.
其中背包容量应设为 mid * x - y + s2 (+ s2 是因为可能买完之后不正好,但价格更低)
还有要各种控制边界,一不小心就会 TLE.
详见代码.
四.代码实现:
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <bitset>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-8
#define lc k * 2
#define rc k * 2 + 1
#define pi acos(-1.0)
#define ll long long int
using namespace std;const int M = (int)1e2;
const ll mod = (ll)1e9 + 7;
const int inf = 0x3f3f3f3f;struct node
{int x, y;int s1, p1;int s2, p2;
}s[M + 5];int dp[(int)6e6 + 5];int n, m;int cal_c(int k, int need)
{int w[2] = {s[k].s1, s[k].s2};int v[2] = {s[k].p1, s[k].p2};dp[0] = 0;int V = need + s[k].s2;for(int i = 1; i <= V; ++i)///这里只能赋初值到 V,不然就T了~dp[i] = inf;for(int i = 0; i < 2; ++i){for(int j = w[i]; j <= V; ++j)dp[j] = min(dp[j], dp[j - w[i]] + v[i]);}int ans = inf;for(int i = need; i <= V; ++i)ans = min(ans, dp[i]);return ans;
}bool check(int mid)
{int c = 0;int num;for(int i = 1; i <= n; ++i){num = mid * s[i].x - s[i].y;if(num <= 0)continue;c += cal_c(i, num);if(c > m)return 0;}return 1;
}int main()
{while(~scanf("%d %d", &n, &m) && (n + m)){for(int i = 1; i <= n; ++i)scanf("%d %d %d %d %d %d", &s[i].x, &s[i].y, &s[i].s1, &s[i].p1, &s[i].s2, &s[i].p2);int l = 0;int r = (int)1e5;while(l < r){int mid = (l + r + 1) >> 1;if(check(mid))l = mid;elser = mid - 1;}printf("%d\n", r);}return 0;
}
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Inviting Friends (HDU
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