第十三届蓝桥杯c++B组国赛 D E

思路: 从高位开始能减成9就先减, 减不了就尽可能地加

范围大用的字符串

#include<iostream>
#include<string>
using namespace std;

string s;
int a, b;
int main() 
{ 
  cin >> s >> a >> b;
  int len = s.size();

  for(int i = 0; i < len; i ++){
    if(b >= s[i]-'0'+1){
      b -= s[i]-'0'+1;
      s[i] = '9';
    }else{
      int d = min(a, 9-s[i]+'0');
      a -= d;
      s[i] = s[i]+d;
    }
  }
  cout << s;
  return 0;
}

Dijkstra

我这里把点权加在了边权的末端, 并让到"1"和到"n"的末端点权为0

没用堆优化,不知道会不会tle

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 1010;
int n, m;
int dist[N], g[N][N], wait[N];
bool st[N];


int dijkstra()
{
  memset(dist, 0x3f, sizeof dist);
  dist[1] = 0;
  
  for(int i = 0; i <= n ; i ++)
  {
    int t = -1;
    for(int j = 1; j <= n; j ++)
      if(!st[j] && (t==-1 || dist[j]<dist[t]))
        t = j;
    
    st[t] = true;
    for(int j = 1; j <= n; j ++){
      dist[j] = min(dist[j], dist[t]+g[t][j]);
    }
  }
  //if(dist[n] == 0x3f3f3f3f) return -1;
  return dist[n];
}
int main()
{
  scanf("%d%d", &n, &m);
  for(int i = 1; i <= n; i ++) scanf("%d", &wait[i]);
  wait[1] = 0;
  wait[n] = 0;
  memset(g, 0x3f, sizeof g);
  
  int a, b, c;
  
  for(int i = 0; i < m; i ++)
  {
    scanf("%d%d%d", &a, &b, &c);
    g[a][b] = c+wait[b];
    g[b][a] = c+wait[a];
  }
  
  printf("%d", dijkstra());
  return 0;
}

总结: 一个多月没练, 寄了