J - Quick Estimates

最新推荐文章于 2025-07-25 13:20:30 发布

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本文介绍了一种帮助工人快速估算家庭维修成本规模的程序设计方案。该方案旨在解决初次访问时成本预估过高或过低的问题，通过计算成本的位数来提供一个大致的成本范围。

Let’s face it… you are not that handy. When you need to make a major home repair, you often need to hire someone to help. When they come for the first visit, they make an estimate of the cost. Here they must be careful: if they overestimate the cost, it might scare you off, but if they underestimate, the work might not be worth their time.

Because the worker is so careful, it can take a long time for them to produce the estimate. But that’s frustrating — when you ask for an estimate, you really are asking for the magnitude of the cost. Will this be $10
or $100 or $1000

? That’s all you really want to know on a first visit.

Please help the worker make the type of estimate you desire. Write a program that, given the worker’s estimate, reports just the magnitude of the cost — the number of digits needed to represent the estimate.
Input

Input begins with a line containing an integer N
(1≤N≤100). The next N lines each contain one estimated cost, which is an integer between 0 and 10100

. (Some of the workers overcharge quite a bit.)
Output

For each estimated cost, output the number of digits required to represent it.

#include <bits/stdc++.h>
using namespace std;
const int maxn=110;
const int inf=0x3f3f3f3f;
int n;
int val[maxn];
int mapp[maxn][maxn];
int dis[maxn],ans[maxn];
bool vis[maxn];
void dij()
{
    for(int i=1; i<=n; i++)
        dis[i]=mapp[1][i];
    dis[1]=0;
    int minn,V=-1,k;
    for (int i = 1; i <=n ; ++i)
    {
        minn=inf;
        k=V;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&minn>dis[j])
            {
                minn=dis[j];
                V=j;
            }

        }
        vis[V]=true;
        for (int j = 1; j <=n ; ++j)
        {
            if(!vis[j]&&dis[j]>dis[V]+mapp[V][j])
            {
                dis[j]=dis[V]+ mapp[V][j];
                ans[j]=ans[V]+val[j];
            }
            else if(!vis[j]&&dis[j]==dis[V]+mapp[V][j])
            {
                ans[j]=max(ans[j],ans[V]+val[j]);
            }
        }
    }

    if(dis[n]==inf)
        printf("impossible\n");
    else
        printf("%d %d\n",dis[n],ans[n]+val[1]);
}
int main()
{
    scanf("%d",&n);
    memset(mapp,0x3f, sizeof(mapp));
    for (int i =1; i <=n ; ++i)
    {
        scanf("%d",&val[i]);
        mapp[i][i]=0;
    }
    int m;
    scanf("%d",&m);
    int x,y,z;

    while(m--)
    {
        scanf("%d%d%d",&x,&y,&z);
        mapp[x][y]=mapp[y][x]=z;
    }
    dij();
    return 0;
}