思路：100个物品的背包问题，其他数值都在1e9
迷之剪枝：维护两个前缀和，如果后面能取完的就直接取完，如果后面能获得的最大价值加上已有的都没有比现在的答案更优那么返回。按照体积由大到小排序，其他排序方法都会T？这是为什么呢
#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
#define LL long long
int n,V;
LL sumw[maxn],sumv[maxn];
struct Node
{
int w,v;
double rate;
}bag[maxn];
bool cmp(Node a,Node b)
{
//return a.rate>b.rate;
return a.v>b.v;
}
LL ans = 0;
void dfs(int pos,int limit,LL get)
{
if(pos==n)
{
if(limit>=bag[pos].v)
ans = max(ans,(LL)get+bag[pos].w);
else
ans = max(ans,get);
return;
}
if(sumv[pos]<=limit)
{
ans = max(ans,get+sumw[pos]);
return;
}
if(get+sumw[pos]<=ans)
return;
dfs(pos+1,limit,get);
if(limit>=bag[pos].v)
dfs(pos+1,limit-bag[pos].v,get+bag[pos].w);
}
int main()
{
while(scanf("%d%d",&n,&V)!=EOF)
{
ans = 0;
sumw[n+1]=0;
sumv[n+1]=0;
for(int i = 1;i<=n;i++)
{
scanf("%d%d",&bag[i].v,&bag[i].w);
bag[i].rate = 1.0*bag[i].w/(double)bag[i].v;
}
sort(bag+1,bag+n+1,cmp);
for(int i = n;i>=1;i--)
{
sumw[i]=sumw[i+1]+bag[i].w;
sumv[i]=sumv[i+1]+bag[i].v;
}
dfs(1,V,0);
printf("%lld\n",ans);
}
}
附上一个0MS的代码
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int n, vLimit, v[100], w[100];
long double rate[100];
long long wBest;
void swap(int i, int j) {
int tmp;
tmp = v[i];
v[i] = v[j];
v[j] = tmp;
tmp = w[i];
w[i] = w[j];
w[j] = tmp;
double tmp2;
tmp2 = rate[i];
rate[i] = rate[j];
rate[j] = tmp2;
}
void sort() {
for (int i = 0; i < n; ++ i) {
for (int j = i + 1; j < n; ++ j) {
if (rate[i] < rate[j]) swap(i, j);
}
}
}
bool can_cut(int k, int _vCurrent, long long _wCurrent) {
long double vCurrent = _vCurrent;
long double wCurrent = _wCurrent;
for (int i = k; i < n && vCurrent < vLimit; ++ i) {
if (vCurrent + v[i] < vLimit) {
vCurrent += v[i];
wCurrent += w[i];
} else {
wCurrent += rate[i] * (vLimit - vCurrent);
vCurrent = vLimit;
}
}
return wCurrent < wBest;
}
void search(int k, int vSum, long long wSum) {
if (wSum > wBest) {
wBest = wSum;
}
if (k < n && !can_cut(k, vSum, wSum)) {
if ((long long)vSum + v[k] <= vLimit) {
search(k + 1, vSum + v[k], wSum + w[k]);
}
search(k + 1, vSum, wSum);
}
}
int main() {
while (scanf("%d%d", &n, &vLimit) != EOF) {
for (int i = 0; i < n; ++ i) {
scanf("%d%d", &v[i], &w[i]);
rate[i] = (long double)w[i] / (long double)v[i];
}
sort();
wBest = 0;
search(0, 0, 0);
std::cout << wBest << std::endl;
}
}
Problem Description
Collecting one's own plants for use as herbal medicines is perhaps one of the most self-empowering things a person can do, as it implies that they have taken the time and effort to learn about the uses and virtues of the plant and how it might benefit them, how to identify it in its native habitat or how to cultivate it in a garden, and how to prepare it as medicine. It also implies that a person has chosen to take responsibility for their own health and well being, rather than entirely surrender that faculty to another. Consider several different herbs. Each of them has a certain time which needs to be gathered, to be prepared and to be processed. Meanwhile a detailed analysis presents scores as evaluations of each herbs. Our time is running out. The only goal is to maximize the sum of scores for herbs which we can get within a limited time.
Input
There are at most ten test cases.
For each case, the first line consists two integers, the total number of different herbs and the time limit.
The
i-th line of the following
n line consists two non-negative integers. The first one is the time we need to gather and prepare the
i-th herb, and the second one is its score.
The total number of different herbs should be no more than
100. All of the other numbers read in are uniform random and should not be more than
109.
Output
For each test case, output an integer as the maximum sum of scores.
Sample Input
3 70 71 100 69 1 1 2
Sample Output
3
