题意:有n个金矿,每个金矿有抓取的消耗的时间和价值,矿工在原点,问在T时间内能得到的最大的价值
分析:唯一和01背包不同的是金矿可能共线,也就是抓取近的金矿后才能抓后面共线的金矿。这是分组背包问题,方法是将点按照斜率排序,如果相等按照距离原点远近排序,将斜率相等的点分成一组,每组的点累加上前面的点的时间和价值,这样每组只选一个点,就是01背包了
收获:分组背包问题
代码:
/************************************************
* Author :Running_Time
* Created Time :2015-8-27 8:48:05
* File Name :J.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
struct Point {
int x, y, t, v;
bool operator < (const Point &r) const {
if (y * r.x == x * r.y) return y < r.y;
else return y * r.x < x * r.y;
}
bool operator == (const Point &r) const {
return y * r.x == x * r.y;
}
}p[N];
int dp[40010];
vector<Point> block[N];
int main(void) {
int n, T, cas = 0;
while (scanf ("%d%d", &n, &T) == 2) {
for (int i=1; i<=n; ++i) {
scanf ("%d%d%d%d", &p[i].x, &p[i].y, &p[i].t, &p[i].v);
}
sort (p+1, p+1+n);
int cnt = 0;
for (int i=1; i<=n; ++i) block[i].clear ();
block[++cnt].push_back (p[1]);
for (int i=2; i<=n; ++i) {
if (p[i] == p[i-1]) block[cnt].push_back (p[i]);
else block[++cnt].push_back (p[i]);
}
for (int i=1; i<=cnt; ++i) {
for (int j=1; j<block[i].size (); ++i) {
block[i][j].t += block[i][j-1].t;
block[i][j].v += block[i][j-1].v;
}
}
memset (dp, 0, sizeof (dp));
for (int i=1; i<=cnt; ++i) {
for (int j=T; j>=0; --j) {
for (int k=0; k<block[i].size (); ++k) {
if (j >= block[i][k].t)
dp[j] = max (dp[j], dp[j-block[i][k].t] + block[i][k].v);
}
}
}
printf ("Case %d: %d\n", ++cas, dp[T]);
}
return 0;
}
原文:http://www.cnblogs.com/Running-Time/p/4765493.html