MZOJ 1344 工作依赖

这道题并不是很难,关键在于读入;

其余只需一个遍历;(考的时候傻逼兮兮的没写出来)

另外,学到了一个 isdigit()用来判断是否是0-9的数字;

 

#include <bits/stdc++.h>
#define read read()
#define up(i,l,r) for(register int i = (l);i <= (r); i++)
#define re(i,u) for(register int i = head[u]; i; i = e[i].nxt) 
using namespace std;
const int N = 10055,M = 1000055;
int n,m,t[N],head[N],ans,size,vis[N];
string s;
struct edge{int v,nxt;}e[M];
void add(int u,int v) {e[++size].v = v; e[size].nxt = head[u]; head[u] = size;}
void dfs(int u)
{    
    ans+=t[u]; vis[u] = 1;
    re(i,u) {int v = e[i].v; if(!vis[v]) dfs(v);}
}
int main()
{
    freopen("work.in","r",stdin);
    scanf("%d%d",&n,&m);
    up(i,1,n)
    {
        int x = 0;
        scanf("%d",&t[i]);
        getline(cin,s);
        up(j,0,(s.size()))
        {
            if(!isdigit(s[j]))
            {
                if(x!=0) add(i,x), x = 0;
                continue;
            }
            else x = x * 10 + s[j] - 48;
        }
    }
    dfs(m);
    printf("%d",ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/mzg1805/p/10316101.html