【JLOI2013】T1、T2、T3 Bzoj3190~3192

T1: 赛车 Bzoj3190

画个s-t图像，写个水平可见直线，注意只维护x>=0的部分。

#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 10005;
int n;
struct line{
	int a,b;int id;
}L[maxn];
bool cmp(line x,line y){return x.ay.b);}
bool ans[maxn];
int Ans[maxn];
stacksta1;stacksta2;
double cross(line x,line y){return (y.b-x.b)*1.0/(x.a-y.a);}
int main(){
	freopen("race.in","r",stdin);
	freopen("race.out","w",stdout);
	scanf("%d",&n);
	int maxb=-100000;
	for(int i=1;i<=n;i++){scanf("%d",&L[i].b);L[i].id=i;maxb=max(maxb,L[i].b);}
	for(int i=1;i<=n;i++){scanf("%d",&L[i].a);if(L[i].b==maxb)ans[i]=1;}
	sort(L+1,L+1+n,cmp);
	sta1.push(0);sta2.push(L[1]);
	for(int i=2;i<=n;i++){
		if(L[i].a==L[i-1].a)continue;
		while(!sta2.empty() && cross(L[i],sta2.top())

T2:卡牌游戏Bzoj3191

是谁不重要，重要的是和庄家的距离，用f[i][j]表示还剩i个人时，从庄家开始数第j个人胜率，枚举所有状态，f[i][j]由f[i-1][*]转移即可

#include
using namespace std;
#define MADOKA main
const int maxn = 55;
int a[maxn];
int n,m;
double f[maxn][maxn];
int MADOKA(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)scanf("%d",&a[i]);
	f[1][1]=1;
	for(int i=2;i<=n;i++)for(int j=1;j<=i;j++)for(int k=1;k<=m;k++){
		int tmp=a[k]%i;
		if(tmp==0)tmp=i;if(tmp==j)continue;
		if(tmp>j)tmp=i+j-tmp;else tmp=j-tmp;
		f[i][j]+=f[i-1][tmp]*1.0/m;
	}
	for(int i=1;i

T3:删除物品Bzoj3192

如图：

#include
using namespace std;
const int maxn = 100005;
#define LL long long
struct node{
	int k,p;
}a[maxn];
int n1,n2,n;
LL ans=0;
bool cmp(node a,node b){
	return a.k>b.k;
}
LL c[maxn];
#define lowbit(x) (x&-x)
void add(int pos,int d){ for(int	i=pos;i<=n;i+=lowbit(i))c[i]+=d; }
LL query(int pos){ LL ret=0;for(int i=pos;i>0;i-=lowbit(i))ret+=c[i];return ret; }
LL query(int pos1,int pos2){ int ret; if(pos1>pos2)ret=0; else ret=query(pos2)-query(pos1-1); return ret; }
int main(){
	freopen("remove.in","r",stdin);
	freopen("remove.out","w",stdout);
	scanf("%d%d",&n1,&n2);
	n=n1+n2;
	for(int i=1;i<=n1;i++){
		scanf("%d",&a[i].k);a[i].p=n1-i+1;
	}
	for(int i=n1+1;i<=n;i++){
		scanf("%d",&a[i].k);a[i].p=i;
	}
	sort(a+1,a+1+n,cmp);
	LL P=n1;
	for(int i=1;i<=n;i++){
		int x=a[i].p;
		if(x < P)ans+=(LL)P-x-query(x+1,P),P=x;
		else if(x>P) ans+=(LL)x-1-P-query(P+1,x-1),P=x-1;
		add(x,1);
	}
	cout<