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  • Codeforces 1039A. Timetable

    题目地址:http://codeforces.com/problemset/problem/1039/A

    题目的关键在于理清楚思路,然后代码就比较容易写了

    对于每一个位置的bus,即对于每一个i(i>=1 && i<=n) ,x[i]必然大于等于 i ,假设第 i 个车可以停在 x[i] 处,则对于j(j>i && j<=x[i]) 令车j停在j-1处,即b[j-1]>=ar[j]+t

    如果x[x[i]]==x[i],只需控制让b[x[i]]<ar[x[i]+1]+t即可

    如果x[x[i]]!=x[i],则x[x[i]]>x[i],则必然有b[x[i]]>=ar[x[i]+1]+t,让x[i]+1个车停在x[i]处,以让x[i]停在x[x[i]]处;由此可以知,i也可以停在x[x[i]]处,与题意相悖,输出No即可

    #include<iostream>
    #include<cstdio> 
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<string.h>
    #include<cstring>
    #include<algorithm>
    #include<set>
    #include<map>
    #include<fstream>
    #include<cstdlib>
    #include<ctime>
    #include<list>
    #include<climits>
    #include<bitset>
    #include<random>
    #include <ctime>
    #include <cassert>
    #include <complex>
    #include <cstring>
    #include <chrono>
    using namespace std;
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("input.in", "r", stdin);freopen("output.in", "w", stdout);
    #define left asfdasdasdfasdfsdfasfsdfasfdas1
    #define tan asfdasdasdfasdfasfdfasfsdfasfdas
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    typedef long long ll;
    typedef unsigned int un;
    const int desll[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
    const int mod=1e9+7;
    const int maxn=2e5+7;
    const int maxm=1e5+7;
    const double eps=1e-4;
    ll m,n;
    ll ar[maxn];
    ll b[maxn];
    int x[maxn];
    int main()
    {
        scanf("%I64d%I64d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%I64d",&ar[i]);
        for(int i=1;i<=n;i++)scanf("%d",&x[i]);
        for(int i=2;i<=n;i++){
            if(x[i]<x[i-1] || x[i]<i){
                printf("No
    ");
                exit(0);
            }
        }
        int i=1;
        while(i<=n)
        {
            int j=i;
            while(j<=n && x[j]==x[i]){
                j++;
            }
            for(int k=i;k<j;k++){
                if(x[k]>k)b[k]=ar[k+1]+m;
                else b[k]=max(b[k-1]+1, ar[k]+m);
                //cout<<k<<" "<<b[k]<<endl;
            }
            if(x[j-1]!=j-1|| (j-1<n && b[j-1]>=ar[j]+m)){
                printf("No
    ");
                exit(0);
            }
            i=j;
        }
        printf("Yes
    ");
        for(ll i=1;i<=n;i++)printf("%I64d%c",b[i],i==n?'
    ':' ');
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wa007/p/9600994.html
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