HDU 4381 - Grid（01背包变形）

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探讨了在给定操作限制下，如何通过最优策略将最多数量的格子从黑色变为白色，并求解达到该状态所需的最少操作次数。

Grid

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 967 Accepted Submission(s): 325

Problem Description

　　There are n boxes in one line numbered 1 to n, at the beginning, all boxes are black. Two kinds of operations are provided to you:

1 a _i x _i ：You can choose any x _i black boxes in interval [1,a _i], and color them white;
2 a _i x _i ：You can choose any x _i black boxes in interval [a _i,n], and color them white;

　　lcq wants to know if she use these operations in optimal strategy, the maximum number of white boxes she can get, and if she get maximum white boxes, the minimum number of operations she should use.
Tips:
1. It is obvious that sometimes you can choose not to use some operations.
2. If the interval of one operation didn’t have enough black boxes, you can’t use this operation.

Input

　　The first line contains one integer T, indicating the number of test case.
　　The first line of each test case contains two integers N (1 <= N <= 1000) and M (1<=M<=1000), indicating that there are N grids and M operations in total. Then M lines followed, each of which contains three integers s _i(1<=s _i<=2) , a _i and x _i (0 <= x _i <= N,1<=a _i<=N), si indicating the type of this operation, a _i and x _i indicating that the interval is [1,a _i] or [a _i,n](depending on s _i), and you can choose x _i black boxes and color them white.

Output

　　For each test case, output case number first. Then output two integers, the first one is the maximum boxes she can get, the second one is the minimum operations she should use.

Sample Input

Sample Output

  
   Case 1: 3 1

Author

WHU

Source

2012 Multi-University Training Contest 9

题意：

有n个点，m个操作k，a,x。

k=1:表明可以在[1,a]内把x个点涂白。

k=2:表明可以在[a,n]内把x个点涂白。

不足x个点不能涂。

我们假设涂的都是连续的，这没有什么影响。对两种操作分别进行01背包。按照端点靠前的优先

得到dpl[x],dpr[x]，代表涂满前x个点最少的操作次数。

然后遍历答案。

#include <string>
#include <string.h>
#include <iostream>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 1000+55;
const int inf = 0x3f3f3f3f;
struct node
{
	int a,x;
}l[maxn],r[maxn];
int dpl[maxn],dpr[maxn];
bool cmd(node a,node b)
{
	return a.a<b.a;
}
int main()
{
	int T,cas=1;
	for(scanf("%d",&T);T--;cas++){
		int n,m;
		scanf("%d %d",&n,&m);
		int lcnt=0,rcnt=0;
		for(int i=1;i<=m;i++){
			int s,a,x;
			scanf("%d %d %d",&s,&a,&x);
			if(s==1){
				l[++lcnt].a=a;
				l[lcnt].x=x;
			}else{
				r[++rcnt].a=n-a+1;
				r[rcnt].x=x;
			}
		}
		sort(l+1,l+1+lcnt,cmd);
		sort(r+1,r+1+rcnt,cmd);
		for(int i=1;i<=n;i++) dpl[i]=dpr[i]=inf;
		dpl[0]=dpr[0]=0;
		for(int i=1;i<=lcnt;i++){
			for(int j=l[i].a;j>=l[i].x;j--){
				dpl[j]=min(dpl[j],dpl[j-l[i].x]+1);
			}
		}
		for(int i=1;i<=rcnt;i++){
			for(int j=r[i].a;j>=r[i].x;j--){
				dpr[j]=min(dpr[j],dpr[j-r[i].x]+1);
			}
		}
		int ansnum=0;
		int ans=0;
		for(int i=0;i<=n;i++){
			for(int j=n-i;j>=0;j--){
				if(dpl[i]+dpr[j]<inf){
					if(i+j>ansnum){
						ansnum=i+j;
						ans=dpl[i]+dpr[j];
						break;
					}else if(i+j==ansnum&&ans>dpl[i]+dpr[j]){
						ans=dpl[i]+dpr[j];
					}
				}
			}
		}
		printf("Case %d: %d %d\n",cas,ansnum,ans);


	}
	return 0;
}