【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<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<stack>
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.a<y.a || (x.a==y.a && x.b>y.b);}
bool ans[maxn];
int Ans[maxn];
stack<double>sta1;stack<line>sta2;
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())<sta1.top())sta1.pop(),sta2.pop();
if(!sta2.empty())sta1.push(cross(L[i],sta2.top()));else sta1.push(0);
sta2.push(L[i]);
}
int cnt=0;
while(!sta2.empty()){ans[sta2.top().id]=1;sta2.pop();}
for(int i=1;i<=n;i++)if(L[i].a==L[i-1].a && L[i].b==L[i-1].b)ans[L[i].id]=ans[L[i-1].id];
int t=0;
for(int i=1;i<=n;i++)if(ans[i])Ans[++t]=i;
printf("%d\n",t);
for(int i=1;i<t;i++)printf("%d ",Ans[i]);printf("%d\n",Ans[t]);
}
/*
4
1 1 0 0
15 16 10 20
*/
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<bits/stdc++.h>
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<n;i++)printf("%.2lf%% ",f[n][i]*100.0);
printf("%.2lf%%\n",f[n][n]*100.0);
}
T3:删除物品Bzoj3192
如图：
#include<bits/stdc++.h>
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<<ans<<endl;
}