2019杭电多校第八场

感觉慢慢陷入了疲态……
依旧是努力签到的一天

1009：Calabash and Landlord

各种分类讨论，交上去，wa，造数据，找到bug，改一改，交上去，wa，造数据，改一改，wa，改一改，wa…循环多次后ac，然后自己也不知道自己写的是个啥了。。
后来听群里老哥们说只要离散化到5*5的格子里然后dfs就好了，明天补补看吧Orz

#include
#define ll long long
using namespace std;
int f(ll x1, ll y1, ll x2, ll y2, ll x, ll y)//1:in  2:out  3:on
{
    if(x > x1 && x < x2 && y > y1 && y < y2) return 1;
    if(x > x2 || x < x1 || y > y2 || y < y1) return 2;
    return 3;
}
int main()
{
	int T;cin>>T;
	while(T--){
        ll x1, y1, x2, y2, x3, y3, x4, y4;
        scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);
        scanf("%lld%lld%lld%lld", &x3, &y3, &x4, &y4);
        if(x1 == x3 && y1 == y3 && x2 == x4 && y2 == y4){
            printf("2\n"); continue;
        }
        if(x1 > x3 || (x1 == x3 && ((y4 < y2 || y3 > y1)))  ){
            swap(x1,x3); swap(y1,y3); swap(x2, x4); swap(y2, y4);
        }
        if(x4 <= x2 && ((y4 >= y2 && y3 <= y1)) ){
            if(x2 == x4){//右对齐
                if(x1 == x3){
                    if(y1 == y3 || y2 == y4) printf("3\n");
                    else printf("4\n");
                    continue;
                }
                else{//x1
                    if(y1 == y3 && y2 == y4){
                        printf("3\n");continue;
                    }
                    else if(y1 == y3 || y2 == y4){
                        printf("4\n"); continue;
                    }
                    else printf("5\n"); continue;
                }
            }
            else if(x1 == x3){
                if(y1 == y3 && y2 == y4){
                    printf("3\n"); continue;
                }
                else if(y1 == y3 || y2 == y4){
                    printf("4\n"); continue;
                }
                else printf("5\n"); continue;
            }
            else if(y1 == y3 && y2 == y4)//x都不相等
            {
                printf("4\n"); continue;
            }
            else if(y1 == y3 || y2 == y4){
                printf("5\n"); continue;
            }
            else {
                printf("6\n"); continue;
            }
        }
        int in = 0, out = 0;
        int t1 = f(x1,y1,x2,y2,x3,y3);
        int t2 = f(x1,y1,x2,y2,x3,y4);
        int t3 = f(x1,y1,x2,y2,x4,y3);
        int t4 = f(x1,y1,x2,y2,x4,y4);
        if(t1 == 1||t2==1||t3==1||t4==1) in = 1;
        if(t1 == 2||t2==2||t3==2||t4==2) out = 1;
        if(in && out) {
            printf("4\n");continue;
        }
        in = 0, out = 0;
        t1 = f(x3, y3, x4, y4, x1, y1);
        t2 = f(x3, y3, x4, y4, x1, y2);
        t3 = f(x3, y3, x4, y4, x2, y1);
        t4 = f(x3, y3, x4, y4, x2, y2);
        if(t1 == 1||t2==1||t3==1||t4==1) in = 1;
        if(t1 == 2||t2==2||t3==2||t4==2) out = 1;
        if(in && out) {
            printf("4\n");continue;
        }
        if(x1 == x3 && x2 == x4){
            if(y1 < y4 && y1 > y3 && y2 > y4) printf("4\n");
            else if(y2 < y4 && y2 > y3 && y1 < y3) printf("4\n");
            else printf("3\n");
        }
        else if(y1 == y3 && y2 == y4){
            if(x1 < x4 && x1 > x3 && x2 > x4) printf("4\n");
            else if(x2 < x4 && x2 > x3 && x1 < x3) printf("4\n");
            else printf("3\n");
        }
        else printf("3\n");
	}
}

1010: Quailty and CCPC

按题意模拟即可。如果用结构体排序的话，不要把名字也存在结构体里面，只要存一个编号就可以了，不然会T。

#include 
#include 
using namespace std;
const int maxn=1e5+5;
struct team
{
	int id;
	int num,pen;
}tea[maxn];
string name[maxn];
bool cmp(team t1,team t2)
{
	if(t1.num==t2.num)return t1.pen<t2.pen;
	else return t1.num>t2.num;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	int T;cin>>T;
	while(T--)
	{
		bool ok=false;
		int n,d;cin>>n>>d;
		for(int i=1;i<=n;i++){
            cin>>name[i]>>tea[i].num>>tea[i].pen;
            tea[i].id = i;
		}
		int ans=n*d;
		if(ans%10!=5)cout<<"Quailty is very great\n";
		else
		{
		    sort(tea+1,tea+1+n,cmp);
		    int id = tea[ans/10+1].id;
			cout<<name[id]<<'\n';
		}
	}
}

1011： Roundgod and Milk Tea

对于每个班级求一下有多少牛奶一定送不出去，这样得到了可以送出去的牛奶数量，和总人数求一个min就是答案。

#include
#define ll long long
using namespace std;
const int maxn = 1e6 + 50;
ll a[maxn], b[maxn];
int main()
{
	int T;cin>>T;
	while(T--){
        int n;scanf("%d",&n);
        ll sum = 0, res = 0;
        ll p = 0;
        for(int i = 0; i < n; ++i){
            scanf("%lld%lld",&a[i],&b[i]);
            sum += b[i];
            p += a[i];
        }
        for(int i = 0; i < n; ++i){
            if(p - a[i] < b[i]) res += (b[i] - (p-a[i]));
        }
        printf("%lld\n", min(sum - res, p));
	}
}