【P1083 借教室】差分数组 + 二分答案

最新推荐文章于 2022-05-03 17:23:30 发布

原创最新推荐文章于 2022-05-03 17:23:30 发布 · 229 阅读

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该博客主要介绍了如何解决P1083题目的教室借用问题，强调了答案的单调性，并提出了采用二分查找的方法。通过构建差分数组并检查前缀和，确定每天实际使用的教室是否满足申请需求。

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P1083
题意就是n天每天有r[i]个教室可用问你有m个申请必须按顺序来从Q[i].l 到 Q[i].r 要用Q[i].d个教室
如果都能满足输出0 如果有按顺序哪个不能满足输出不能满足
我们首先发现答案继续单调性那么可以二分
二分的时候check数组如何满足检验呢？
其实就是差分数组加一下前缀和求每天的实际使用教室值和最大教室判断一下check即可
check函数为此

bool check(int mid)
{
    memset(b,0,sizeof(b));
    ll tmp = 0;
    for(int i = 1;i<=mid;++i)
        b[Q[i].l]+=Q[i].d,b[Q[i].r+1]-=Q[i].d;
    sum[0] = 0;
    for(int i = 1;i<=n;i++)
    {
        sum[i] = sum[i-1]+b[i];
        if(sum[i]>a[i]) return false;
    }
    return true;
}

代码如下

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 1000025;
int n,m;
ll a[MAX_N],b[MAX_N],sum[MAX_N];
struct node
{
    ll d,l,r;
}Q[MAX_N];
bool check(int mid)
{
    memset(b,0,sizeof(b));
    ll tmp = 0;
    for(int i = 1;i<=mid;++i)
        b[Q[i].l]+=Q[i].d,b[Q[i].r+1]-=Q[i].d;
    sum[0] = 0;
    for(int i = 1;i<=n;i++)
    {
        sum[i] = sum[i-1]+b[i];
        if(sum[i]>a[i]) return false;
    }
    return true;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i = 1;i<=n;++i) scanf("%lld",&a[i]);
    for(int i = 1;i<=m;++i)
    {
        scanf("%lld%lld%lld",&Q[i].d,&Q[i].l,&Q[i].r);
    }
    int l = 1,r = m;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        if(check(mid)) l=mid+1;
        else r = mid - 1;
    }
    if(r>=m) printf("0\n");
    else
    {
        printf("-1\n%d\n",r+1);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}