uva12657 - Boxes in a Line 入门经典II 第六章数据结构基础 例题6-5

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=602&page=show_problem&problem=4395

说明：按着题中的思路敲的，本题还是有不少收获。

自己敲的代码：

<span style="font-size:18px;">#include<stdio.h>
//#include<iostream>//这里不能用iostream，和right，left冲突
#include<algorithm>
using namespace std;
const int maxn= 100000 + 10;
int left[maxn],right[maxn];

inline void link(int L, int R) {
  right[L] = R; left[R] = L;
}

int main(){
    int n,m,kase=1;
    while(scanf("%d%d",&n,&m)!=EOF){//用cin习惯了，没有加EOF，TLE了。
        for(int i=0;i<n;i++){
            link(i,i+1);
        }
        left[0]=n;right[n]=0;

        int op,inv=0,x,y;
        while(m--){
            scanf("%d",&op);
            if(op==4) {inv=!inv;continue;}
            scanf("%d%d",&x,&y);

            if(op==3 && right[y]==x) swap(x,y);
            if(op!=3 && inv) op=3-op;
            if(op==1 && left[y]==x) continue;
            if(op==2 && right[y]==x) continue;

            int lx=left[x],rx=right[x],ly=left[y],ry=right[y];
            if(op==1){
                link(lx,rx);link(ly,x);link(x,y);
            }else if(op==2){
                link(lx,rx);link(y,x);link(x,ry);
            }else if(op==3){
                if(right[x]==y) {
                    link(lx,y);link(y,x);link(x,ry);
                }else{
                    link(lx,y);link(y,rx);
                    link(ly,x);link(x,ry);
                }
            }
        }

        long long sum=0;
        int b=0;
        for(int i=1;i<=n;i++){
            b=right[b];
            if(i%2) sum+=b;
        }
        if(inv && n%2==0) sum=(long long)n*(n+1)/2-sum;
        printf("Case %d: %lld\n",kase++,sum);
    }
    return 0;
}
</span>

汝佳提供的代码：

// UVa12657 Boxes in a Line
// Rujia Liu
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 100000 + 5;
int n, left[maxn], right[maxn];

inline void link(int L, int R) {
  right[L] = R; left[R] = L;
}

int main() {
  int m, kase = 0;
  while(scanf("%d%d", &n, &m) == 2) {
    for(int i = 1; i <= n; i++) {
      left[i] = i-1;
      right[i] = (i+1) % (n+1);
    }
    right[0] = 1; left[0] = n;
    int op, X, Y, inv = 0;

    while(m--) {
      scanf("%d", &op);
      if(op == 4) inv = !inv;
      else {
        scanf("%d%d", &X, &Y);
        if(op == 3 && right[Y] == X) swap(X, Y);
        if(op != 3 && inv) op = 3 - op;
        if(op == 1 && X == left[Y]) continue;
        if(op == 2 && X == right[Y]) continue;

        int LX = left[X], RX = right[X], LY = left[Y], RY = right[Y];
        if(op == 1) {
          link(LX, RX); link(LY, X); link(X, Y);
        }
        else if(op == 2) {
          link(LX, RX); link(Y, X); link(X, RY);
        }
        else if(op == 3) {
          if(right[X] == Y) { link(LX, Y); link(Y, X); link(X, RY); }
          else { link(LX, Y); link(Y, RX); link(LY, X); link(X, RY); }
        }
      }
    }

    int b = 0;
    long long ans = 0;
    for(int i = 1; i <= n; i++) {
      b = right[b];
      if(i % 2 == 1) ans += b;
    }
    if(inv && n % 2 == 0) ans = (long long)n*(n+1)/2 - ans;
    printf("Case %d: %lld\n", ++kase, ans);
  }
  return 0;
}