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NC14302. Confliction

描述

Recently, Alice and Bob are working on a resource-sharing computation model. In this model, there are two processing units, and a memory space which could be represented as unlimited linear grids ..., A-3, A-2, A-1, A0, A1, A2, A3, ... Each processing units has a pointer to mark exactly one grid in the memory. In each clock turn, the pointer would stay at the current grid, move the pointer one grid forward, or move the pointer one grid backward. In their work, Alice and Bob would submit their codes, and their programs would start at the same time. Initially, both pointers would be located at a random grid, and move according to a set of instructions. If both pointers are at the same grid at the same time, the confliction counter would plus one and record it(If their initial grids are the same, the counter would still record it). Now it is your job to find the maximum conflictions the counter could record.

输入描述

The first line is the number of test cases.
For each test case, the first is an integer NAlice(NAlice ≤100000), donating the length of the instructions of Alice.
The next NAlice lines describe Alice`s instructions. Each line consists of two integer c, t. c could be -1, 0, 1, donating moving forward, staying, and moving backward respectively. t is a non-negative integer donating that the instruction c would be executed t times, in the next t clock turns.
The next line is an integer NBob (NBob ≤ 100000), donating the length of the instructions of Bob.
The next NBob lines describe Bob`s instructions. Each line consists of two integer c, t. c could be -1, 0, 1, donating moving forward, staying, and moving backward respectively. t is a non-negative integer donating that the instruction c would be executed t times, in the next t clock turns.
Suppose LAlice equals the sum of all t in Alice's program, and LBob equals the sum of all t in Bob's program.
It is guaranteed LAlice = LBob and LAlice, LBob ≤ 1018.

输出描述

For each test case, output a line containing the maximum conflictions.

示例1

输入: 

2
1
1 2
2
1 1
-1 1
1
0 6
4
-1 2
1 1
-1 2
1 1

输出: 

2
3

原站题解

上次编辑到这里，代码来自缓存 点击恢复默认模板

C++11(clang++ 3.9) 解法, 执行用时: 947ms, 内存消耗: 12032K, 提交时间: 2017-11-11 18:04:33 

#include<cstdio>
#include<algorithm>
using namespace std;
#define LL long long
struct Node{
	LL l,t;
	int dir,flag;
	bool operator < (const Node& c)const{
		return l<c.l||(l==c.l&&flag==1&&c.flag==-1);
	}
}c[400005];
LL a[100001],b[100001];
int da[100001],db[100001];
int main()
{
	int test,n,m;
	scanf("%d",&test);
	while(test--){
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%d%lld",&da[i],&a[i]);
		da[n]=0;a[n]=1;
		scanf("%d",&m);
		for(int i=0;i<m;i++)
			scanf("%d%lld",&db[i],&b[i]);
		db[m]=0;b[m]=1;
		LL pos=0;
		int cnt=0;
		for(int i=0,j=0;i<=n&&j<=m;){
			LL tmp=min(a[i],b[j]);
			int dir=da[i]-db[j];
			LL d=pos+dir*tmp;
			LL l=min(pos,d-dir);
			c[cnt++]={l,tmp,abs(dir),1};
			c[cnt++]={l+(tmp-1)*abs(dir),tmp,abs(dir),-1};
			a[i]-=tmp;b[j]-=tmp;
			if(!a[i])i++;
			if(!b[j])j++;
			pos=d;
		}
		sort(c,c+cnt);
		LL ans=0,num[2]={0,0};
		for(int i=0;i<cnt;i++){
			if(c[i].dir==0)num[c[i].l&1]+=c[i].t*c[i].flag;
			else if(c[i].dir==1){num[0]+=c[i].flag;num[1]+=c[i].flag;}
			else if(c[i].dir==2)num[c[i].l&1]+=c[i].flag;
			ans=max(ans,num[c[i].l&1]);
			if(i+1<cnt&&c[i].l+1<c[i+1].l)ans=max(ans,num[~c[i].l&1]);
		}
		printf("%lld\n",ans);
	}
}

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