题解：P13316 [GCJ 2012 #1A] Kingdom Rush

#include <bits/stdc++.h>
using namespace std;
const int N=1e3+5;
int t,sum,w[N];
struct one{
	int _1,_2,id;
}l1[N];
struct two{
	int _2,id;
}l2[N];

bool cmp1(one x,one y){
	if(sum>=x._1&&sum>=y._1)	return x._2>y._2;
	return x._1<y._1;
}//1星
bool cmp2(two x,two y){
	return x._2<y._2;
}//2星
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
    cin>>t;
    int k=0;
    while(t--){
    	int n,ans=0;
        sum=0;
    	cin>>n;
        memset(w,0,sizeof(w));//清零
    	for(int i=1;i<=n;i++){
    		cin>>l1[i]._1>>l2[i]._2;
    		l1[i]._2=l2[i]._2;
    		l1[i].id=i;
    		l2[i].id=i;
		}
		sort(l1+1,l1+1+n,cmp1);//先排1星
		sort(l2+1,l2+1+n,cmp2);//再排2星

		int x1=1,x2=1;//开贪
		bool f=0;
		while(x2<=n){
			bool _f=0;
			if(sum>=l2[x2]._2){
				sum+=(2-w[l2[x2].id]);
                w[l2[x2].id]=2;
                _f=1;
				x2++;
				ans++;
			}else{
				if(sum>=l1[x1]._1&&x1<=n){//这里卡了好久，一定要判断x1值域，不然会爆TAT
        			if(w[l1[x1].id]==0){
                        w[l1[x1].id]=1;
						sum++;
						ans++;
					}
					x1++;
					_f=1;
				}
			}
            sort(l1+x1,l1+1+n,cmp1);//重新排
			if(!_f){
				cout<<"Case #"<<++k<<": Too Bad\n";
				f=1;
				break;
			}//此关卡无法通过
		}
		if(f)	continue;
		cout<<"Case #"<<++k<<": "<<ans<<"\n";//此关卡可通过
    }
    return 0;//完结撒花(OvO)
}