nyoj 248 BUYING FEED(欲張りorDP)
4937 ワード
時間制限:
3000 ms|メモリ制限:
65535 KB
難易度:4
説明
Farmer John needs to travel to town to pick up K (1 <= K <= 100)pounds of feed. Driving D miles with K pounds of feed in his truck costs D*K cents.
The county feed lot has N (1 <= N<= 100) stores (conveniently numbered 1..N) that sell feed. Each store is located on a segment of the X axis whose length is E (1 <= E <= 350). Store i is at location X_i (0 < X_i < E) on the number line and can sell John as much as F_i (1 <= F_i <= 100) pounds of feed at a cost of C_i (1 <= C_i <= 1,000,000) cents per pound. Amazingly, a given point on the X axis might have more than one store.
Farmer John starts at location 0 on this number line and can drive only in the positive direction, ultimately arriving at location E, with at least K pounds of feed. He can stop at any of the feed stores along the way and buy any amount of feed up to the the store's limit. What is the minimum amount Farmer John has to pay to buy and transport the K pounds of feed? Farmer John knows there is a solution. Consider a sample where Farmer John needs two pounds of feed from three stores (locations: 1, 3, and 4) on a number line whose range is 0..5: 0 1 2 3 4 5 --------------------------------- 1 1 1 Available pounds of feed 1 2 2 Cents per pound
It is best for John to buy one pound of feed from both the second and third stores. He must pay two cents to buy each pound of feed for a total cost of 4. When John travels from 3 to 4 he is moving 1 unit of length and he has 1 pound of feed so he must pay1*1 = 1 cents.
When John travels from 4 to 5 heis moving one unit and he has 2 pounds of feed so he must pay 1*2 = 2 cents. The total cost is 4+1+2 = 7 cents.
入力
The first line of input contains a number c giving the number of cases that follow
There are multi test cases ending with EOF.
Each case starts with a line containing three space-separated integers: K, E, and N
Then N lines follow :every line contains three space-separated integers: Xi Fi Ci
しゅつりょく
For each case,Output A single integer that is the minimum cost for FJ to buy and transport the feed
サンプル入力
1
2 5 3
3 1 2
4 1 2
1 1 1
サンプル 7
ソース
3 プログラム
アップロード
ACM_
にやった は があったので、この はDPだと っていました. は ることができると きました.
いきなり たのにまだ すぎて... も らなかったわけじゃないただ く えていただけ
りAになってからは が に っていたのは いないと じたのでコメントエリアを てやっと のお が じポイントにあるかもしれないことを りました........................................................................................................................ を いた.の
DPでもやった
る :
+ある から までの を さいものから きいものに べ えるとよい
ったのは いに いない.
コード:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
int x;
int w;
int per;
}c[400];
int n,e,k;
bool cmp(node a,node b)
{
return n-a.x+a.per<n-b.x+b.per;
}
int main()
{
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
memset(&c,0,sizeof(&c));
scanf("%d %d %d",&k,&e,&n);
for(int i=0;i<n;i++)
{
scanf("%d %d %d",&c[i].x,&c[i].w,&c[i].per);
}
sort(c,c+n,cmp);
int cost=0;
for(int i=0;i<n;i++)
{
if(k>=c[i].w)
{
cost+=(e-c[i].x)*c[i].w+c[i].w*c[i].per;
k-=c[i].w;
}
else
{
cost+=(e-c[i].x)*k+k*c[i].per;
k=0;
}
if(k==0)
break;
}
printf("%d
",cost);
}
return 0;
}
DP :
dp[x][w]xが の wである み
for i=0.....FI
dp[x][w]=min(dp[x][w],dp[x-1][w-i]+i*Ci+(e-x+1)*i);
ACコード:
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <string.h>
using namespace std;
struct node
{
int w;
int per;
};
int main()
{
int k,e,n;
int dp[400][105];
int ncase;
vector<node>edge[400];
scanf("%d",&ncase);
while(ncase--)
{
memset(dp,100,sizeof(dp));
memset(edge,0,sizeof(edge));
scanf("%d %d %d",&k,&e,&n);
for(int i=0;i<n;i++)
{
int xi;
node temp;
scanf("%d %d %d",&xi,&temp.w,&temp.per);
edge[xi].push_back(temp);
}
dp[0][0]=0;
for(int i=1;i<=e;i++)
{
for(int l=k;l>=0;l--)
{
dp[i][l]=dp[i-1][l];
for(int j=0;j<edge[i-1].size();j++)
{
int Per=edge[i-1][j].per;
int W=edge[i-1][j].w;
for(int p=0;p<=W;p++)
if(l>=p)
dp[i][l]=min(dp[i][l],dp[i-1][l-p]+p*Per+(e-i+1)*p);
}
}
}
printf("%d
",dp[e][k]);
}
return 0;
}