NOIP2014
DAY1
生活大爆炸版石头剪刀布
直接模拟即可。
// codevs3716
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define MOD 1000000007
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int maxn=2000;
int c[5][5]={{0,0,1,1,0},
{1,0,0,1,0},
{0,1,0,0,1},
{0,0,1,0,1},
{1,1,0,0,0}};
int a[maxn],b[maxn];
int main() {
int n,na,nb;
scanf("%d%d%d",&n,&na,&nb);
for (int i=0;i<na;i++) scanf("%d",&a[i]);
for (int i=0;i<nb;i++) scanf("%d",&b[i]);
int A=0,B=0;
for (int i=0;i<n;i++) {
A+=c[a[i%na]][b[i%nb]];
B+=c[b[i%nb]][a[i%na]];
}
printf("%d %d",A,B);
return 0;
}
联合权值
一开始无脑枚举，因为只需要dfs1层，感觉完全不虚，结果被菊花树卡得只有70分，于是怒水一发树形dp。
// codevs3728
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define MOD 10007
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int maxn=200010;
struct edge {int to,next;}e[maxn<<1];
int head[maxn],n,cnt,w[maxn],ans1,ans2;;
void link(int u,int v) {
e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;
e[++cnt].to=u;e[cnt].next=head[v];head[v]=cnt;
}
void dfs(int x,int fa,int f) {
ans1=(ans1+w[x]*w[f]);ans2=max(ans2,w[x]*w[f]);
int x1=0,x2=0;
for (int i=head[x];i;i=e[i].next) if (e[i].to!=fa) {
dfs(e[i].to,x,fa);
ans2=max(ans2,x2*w[e[i].to]);
ans1=(ans1+x1*w[e[i].to])%MOD;
x1=(x1+w[e[i].to])%MOD;x2=max(x2,w[e[i].to]);
}
}
int main() {
scanf("%d",&n);
for (int u,v,i=1;i<n;i++) {
scanf("%d%d",&u,&v);
link(u,v);
}
for (int i=1;i<=n;i++) scanf("%d",&w[i]);
dfs(1,0,0);
printf("%d %d",ans2,ans1*2%MOD);
return 0;
}
飞扬的小鸟
一开始打了个nm²加队列乱搞可以获得85分的高分哦！正解背包。
// codevs3729
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 100000000
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int maxn=10010,maxm=1010;
struct data {int p,l,r;}t[maxn];
int n,m,K,u[maxn],d[maxn],l[maxn],r[maxn],f[maxn][maxm];
int main() {
scanf("%d%d%d",&n,&m,&K);
for (int i=0;i<n;i++) scanf("%d%d",&u[i],&d[i]);
for (int i=0;i<=n;i++) l[i]=0,r[i]=m+1;
for (int x,i=1;i<=K;i++) {
scanf("%d",&x);
scanf("%d%d",&l[x],&r[x]);
}
int cnt=0;
for (int i=1;i<=n;i++) {
for (int j=1;j<=m;j++) {
f[i][j]=inf;
if (j>u[i-1]) f[i][j]=min(f[i][j],min(f[i-1][j-u[i-1]],f[i][j-u[i-1]])+1);
}
for (int j=m-u[i-1];j<=m;j++) f[i][m]=min(f[i][m],min(f[i-1][j],f[i][j])+1);
for (int j=l[i]+1;j<=r[i]-1;j++)
if (j+d[i-1]<=m) f[i][j]=min(f[i][j],f[i-1][j+d[i-1]]);
for (int j=1;j<=l[i];j++) f[i][j]=inf;
for (int j=r[i];j<=m;j++) f[i][j]=inf;
int flag=0;
for (int j=1;j<=m;j++) if (f[i][j]<inf) {flag=1;break;}
if (!flag) {printf("0\n%d",cnt);return 0;}
else if (r[i]!=m+1) cnt++;
}
int ans=inf;
for (int i=1;i<=m;i++) ans=min(ans,f[n][i]);
printf("1\n%d",ans);
return 0;
}
DAY2
无线网络发射选址
无脑枚举。
// codevs3730
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define MOD 10007
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
int f[200][200],n,d;
int main() {
scanf("%d%d",&d,&n);
for (int x,y,k,i=1;i<=n;i++) {
scanf("%d%d%d",&x,&y,&k);
f[x][y]=k;
}
int ans=0,tot=0;
for (int i=0;i<=128;i++)
for (int j=0;j<=128;j++) {
int cnt=0;
for (int k=max(i-d,0);k<=min(i+d,128);k++)
for (int l=max(j-d,0);l<=min(j+d,128);l++) cnt+=f[k][l];
if (ans==cnt) tot++;
else if (ans<cnt) ans=cnt,tot=1;
}
printf("%d %d",tot,ans);
return 0;
}
寻找道路
写得奇丑无比。。还Wa了两发，数组开小了→_→。。先反向连边处理哪些点能走哪些点不能走，然后Dijkstra。
// codevs3731
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define MOD 10007
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int maxn=10010,maxm=200010;
struct edge {int to,next;}e[maxm<<1];
struct data {
int num,w;
friend bool operator < (const data a,const data b) {
return a.w>b.w;
}
};
int head[maxn],f[maxn],vis[maxn],dis[maxn],u[maxm],v[maxm];
int n,m,cnt,s,t;
void link(int u,int v) {
e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;
}
void dfs(int x) {
vis[x]=1;
for (int i=head[x];i;i=e[i].next) if (!vis[e[i].to]) dfs(e[i].to);
}
void Dijkstra() {
priority_queue<data> q;
data x=(data){s,0},y;
for (int i=1;i<=n;i++) dis[i]=inf,vis[i]=0;
dis[s]=0;q.push(x);
while (!q.empty() && !vis[t]) {
x=q.top();q.pop();
if (vis[x.num]) continue;
vis[x.num]=1;
for (int i=head[x.num];i;i=e[i].next)
if (f[e[i].to] && dis[e[i].to]>x.w+1) {
dis[e[i].to]=y.w=x.w+1;
y.num=e[i].to;
q.push(y);
}
}
}
int main() {
scanf("%d%d",&n,&m);
for (int i=1;i<=m;i++) {
scanf("%d%d",&u[i],&v[i]);
link(v[i],u[i]);
//link(u,v);
}
scanf("%d%d",&s,&t);swap(s,t);
dfs(s);
memset(head,0,sizeof(head));cnt=0;
for (int i=1;i<=m;i++) link(u[i],v[i]);
for (int i=1;i<=n;i++) {
f[i]=vis[i];
for (int j=head[i];j;j=e[j].next) f[i]&=vis[e[j].to];
}
swap(s,t);
Dijkstra();
printf("%d",dis[t]==inf ? -1 : dis[t]);
return 0;
}
解方程
一开始一直纠结怎么优化高精度，一直无果。。模了题解没想到是这这样的结果→_→。
我们发现若将等式左侧模上一个数等于0，那么有可能这个x是解。而在模M的意义下,f[x]与f[x+M]的值是一样的。于是我们就随便搞5个素数，分别预处理出从1~M-1的范围中的解，然后枚举x判断即可。
// codevs3732
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define MOD 10007
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int maxn=1000010;
int M[5]={9973,9931,9941,9949,9967};
int ans[maxn],a[5][maxn],res[5][maxn],pre[5][maxn];
int n,m;
int cal(int t,int x) {
int sum=0;
for (int i=0;i<=n;i++) sum=(sum+a[t][i]*pre[t][i])%M[t];
if (sum<0) sum+=M[t];
return sum;
}
bool check(int x) {
for (int t=0;t<5;t++) if (res[t][x%M[t]]!=0) return 0;
return 1;
}
int main() {
scanf("%d%d",&n,&m);
char ch[10010];
for (int i=0;i<=n;i++) {
scanf("%s",ch+1);
int l=strlen(ch+1);
bool flag=0;
for (int t=0;t<5;t++) {
if (ch[1]!='-') a[t][i]=ch[1]-'0';
else a[t][i]=0,flag=1;
}
for (int t=0;t<5;t++) {
for (int k=2;k<=l;k++) a[t][i]=(a[t][i]*10+ch[k]-'0')%M[t];
if (flag) a[t][i]=-a[t][i];
}
}
for (int t=0;t<5;t++)
for (int x=1;x<M[t];x++) {
pre[t][0]=1;
for (int i=1;i<=n;i++) pre[t][i]=(pre[t][i-1]*x)%M[t];
res[t][x]=cal(t,x);
}
for (int i=1;i<=m;i++) if (check(i)) ans[++ans[0]]=i;
printf("%d\n",ans[0]);
for (int i=1;i<=ans[0];i++) printf("%d\n",ans[i]);
return 0;
}
This passage is made by MashiroSky.
