# [BZOJ1649][Usaco2006 Dec]Cow Roller Coaster

## 1649: [Usaco2006 Dec]Cow Roller Coaster

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 714  Solved: 361 [Submit][Status][Discuss]

## Description

The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget. The roller coaster will be built on a long linear stretch of land of length L (1 <= L <= 1,000). The roller coaster comprises a collection of some of the N (1 <= N <= 10,000) different interchangable components. Each component i has a fixed length Wi (1 <= Wi <= L). Due to varying terrain, each component i can be only built starting at location Xi (0 <= Xi <= L-Wi). The cows want to string together various roller coaster components starting at 0 and ending at L so that the end of each component (except the last) is the start of the next component. Each component i has a "fun rating" Fi (1 <= Fi <= 1,000,000) and a cost Ci (1 <= Ci <= 1000). The total fun of the roller coster is the sum of the fun from each component used; the total cost is likewise the sum of the costs of each component used. The cows' total budget is B (1 <= B <= 1000). Help the cows determine the most fun roller coaster that they can build with their budget.

## Input

* Line 1: Three space-separated integers: L, N and B.

* Lines 2..N+1: Line i+1 contains four space-separated integers, respectively: Xi, Wi, Fi, and Ci.

第1行输入L，N，B，接下来N行，每行四个整数Xi，wi，Fi，Ci．

## Output

* Line 1: A single integer that is the maximum fun value that a roller-coaster can have while staying within the budget and meeting all the other constraints. If it is not possible to build a roller-coaster within budget, output -1.

5 6 10
0 2 20 6
2 3 5 6
0 1 2 1
1 1 1 3
1 2 5 4
3 2 10 2

## Sample Output

17

I'm 安轨
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char buf[10000000], *ptr = buf - 1;
int f = 1, n = 0;
char ch = *++ptr;
while(ch < '0' || ch > '9'){
if(ch == '-') f = -1;
ch = *++ptr;
}
while(ch <= '9' && ch >= '0'){
n = (n << 1) + (n << 3) + ch - '0';
ch = *++ptr;
}
return f * n;
}
struct Node{
int x, w, f, c;
bool operator < (const Node &a) const {
return x < a.x;
}
}a[10000 + 10];
int L, N, B;
int dp[1000 + 10][1000 + 10];
int main(){
memset(dp, -1, sizeof(dp));
for(int i = 1; i <= N; i++){
}
sort(a + 1, a + N + 1);
dp[0][0] = 0;
for(int i = 1; i <= N; i++){
for(int j = 0; j + a[i].c <= B; j++)
if(dp[a[i].x][j] != -1)
dp[a[i].x + a[i].w][j + a[i].c] = max(dp[a[i].x + a[i].w][j + a[i].c], dp[a[i].x][j] + a[i].f);
}
int ans = -1;
for(int i = 0; i <= B; i++)
ans = max(ans, dp[L][i]);
printf("%d\n", ans);
return 0;
}

posted @ 2017-09-25 20:14  Elder_Giang  阅读(107)  评论(0编辑  收藏  举报