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Spreadsheet Tracking
紫书第二个思路十分巧妙，能用很少的代码解出此题。
AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
int t1;
struct node
{
string idx;
int n;
int arr[15];
}a[100];
int dx,dy;
bool ju;
void move(int x,int y)
{
rep(i,1,t1) {
int dl=0,dr=0,il=0,ir=0;
if(a[i].idx=="DR") {
rep(j,1,a[i].n) {
if(a[i].arr[j]==x) {ju=true;return ;}
if(a[i].arr[j]<x) dr++;
}
}
else if(a[i].idx=="DC") {
rep(j,1,a[i].n) {
if(a[i].arr[j]==y) {ju=true;return ;}
if(a[i].arr[j]<y) dl++;
}
}
else if(a[i].idx=="IR") {
rep(j,1,a[i].n) {
if(a[i].arr[j]<=x) ir++;
}
}
else if(a[i].idx=="IC") {
rep(j,1,a[i].n)
if(a[i].arr[j]<=y) il++;
}
else if(a[i].idx=="EX") {
if(a[i].arr[1]==x&&a[i].arr[2]==y) {x=a[i].arr[3];y=a[i].arr[4];}
else if(a[i].arr[3]==x&&a[i].arr[4]==y) {x=a[i].arr[1];y=a[i].arr[2];}
}
x=x-dr;y=y-dl;x=x+ir;y=y+il;
/*cout<<x<<" "<<y<<endl;*/
}
dx=x;dy=y;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
//freopen("out.txt", "w", stdout);
ios::sync_with_stdio(0),cin.tie(0);
int l,r;
int cnt=0;
while(cin>>l>>r&&l) {
cnt++;
if(cnt!=1)
cout<<endl;
cin>>t1;
rep(i,1,t1) {
cin>>a[i].idx;
if(a[i].idx=="EX") cin>>a[i].arr[1]>>a[i].arr[2]>>a[i].arr[3]>>a[i].arr[4];
else {cin>>a[i].n;rep(j,1,a[i].n) cin>>a[i].arr[j];}
}
int t2;
cin>>t2;
cout<<"Spreadsheet #"<<cnt<<endl;
rep(i,1,t2) {
int x,y;
dx=0;dy=0;ju=false;
cin>>x>>y;
move(x,y);
if(ju) cout<<"Cell data in ("<<x<<","<<y<<") GONE"<<endl;
else cout<<"Cell data in ("<<x<<","<<y<<") moved to ("<<dx<<","<<dy<<")"<<endl;
}
}
return 0;
}