UVA-12657 移动盒子 题解答案代码 算法竞赛入门经典第二版

​​GitHub - jzplp/aoapc-UVA-Answer: 算法竞赛入门经典 例题和习题答案 刘汝佳 第二版​​

大致方法和算法竞赛入门经典中一样。

AC代码

#include<stdio.h>

#define MAXN 100005
int left[MAXN];
int right[MAXN];

int inv, n, m;

void init() {
  int i;
  inv = 0;
  for(i = 0; i <= n; ++i) { 
    left[i] = (i + n) % (n + 1);
    right[i] = (i + 1) % (n + 1);
  }
}

void swap(int x, int y) {
  int xl = left[x], xr = right[x];
  int yl = left[y], yr = right[y];
  
  if(right[x] != y && right[y] != x) {
    right[xl] = y;
    left[y] = xl;
    
    left[xr] = y;
    right[y] = xr;
    
    right[yl] = x;
    left[x] = yl;
    
    left[yr] = x;
    right[x] = yr;
    return;
  }
  
  if(right[x] == y) {
    right[xl] = y;
    left[y] = xl;
    
    left[yr] = x;
    right[x] = yr;
    
    left[x] = y;
    right[y] = x;
    return;
  }
  
  if(right[y] == x) {
    right[yl] = x;
    left[x] = yl;
    
    left[xr] = y;
    right[y] = xr;
    
    left[y] = x;
    right[x] = y;
    return;
  }
}

void change1(int x, int y) {
  if(right[x] == y) {
    return;
  }
  int xl = left[x], xr = right[x];
  int yl = left[y], yr = right[y];
  
  right[xl] = xr;
  left[xr] = xl;
  
  right[yl] = x;
  left[x] = yl;
  
  left[y] = x;
  right[x] = y;
}

void change2(int x, int y) {
  if(left[x] == y) {
    return;
  }
  int xl = left[x], xr = right[x];
  int yl = left[y], yr = right[y];
  
  right[xl] = xr;
  left[xr] = xl;
  
  left[yr] = x;
  right[x] = yr;
  
  right[y] = x;
  left[x] = y;
}

long long count() {
  int i;
  int type, x, y;
  for(i = 0; i < m; ++i) {
    scanf("%d", &type);
    if(type < 4) {
      scanf("%d %d", &x, &y);
    } else {
      inv = 1 - inv;
      continue;
    }
    if(type == 3) {
      swap(x, y);
      continue;
    }
    if(inv) {
      type = 3 - type;
    }
    if(type == 1) {
      change1(x, y);
      continue;
    }
    if(type == 2) {
      change2(x, y);
      continue;
    }
  }
  
  x = 0;
  long long sum = 0;
  for(i = 1; i <= n; ++i) {
    x = right[x];
    if(i % 2 == 1)
      sum += x;
  }
  if(inv && n % 2 == 0) {
    sum = n * (long long)(n + 1) / 2 - sum;
  }
  return sum;
}


int main() {
  int i = 0;
  long long sum = 0;
  while(scanf("%d %d", &n, &m) == 2) {
    ++i;
    init();
    sum = count();
    printf("Case %d: %lld\n", i, sum);
  }
  return 0;
}