整理第十届河南省ACM省赛正解

问题A：谍报分析

题目链接

正解：

#include<stdio.h>
#include<string.h>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
struct Word
{
char str[100];
int count;
}word[1000];
int N=0;
int comp(Word w1,Word w2)
{
if(w1.count!=w2.count)
return w1.count>w2.count;
else
{
int a=strcmp(w1.str,w2.str);
if(a>0)
return 0;
else
return 1;
}
}
int main()
{
map<string,int>m;
map<string,int>::iterator it;
char str[100];
while(scanf("%s",str)!=EOF)
{
int len=strlen(str);
if(str[len-1]==','||str[len-1]=='.')
str[len-1]='\0';
string s=str;
m[s]++;
}
for(it=m.begin();it!=m.end();it++)
{
string s=it->first;
int count=it->second;
strcpy(word
.str,s.c_str());
word[N++].count=count;
}
sort(word,word+N,comp);
for(int i=0;i<10;i++)
{
printf("%s %d\n",word[i].str,word[i].count);
}
return 0;
}

B：情报传递

题目链接

正解：

#include<iostream>
#include<vector>
#include<queue>
#include<stdio.h>
#include<sstream>
using namespace std;
int n, m;
int up[5100];
int build[5100];
vector<int> down[5100];
int Send(int s)
{
int ret = 0;
// 顺着s一直找上级，一直找到0
while (s)
{
if (!build[s])
{
build[s] =true;
ret++;
}
s = up[s];
}
// 注意0本身也要向上建立通道。
// 因为样例的第一行 Send 0  输出为 0
if (!build[0])
{
build[0] = true;
ret++;
}
return ret;
}

int Danger(int s)
{
int ret = 0;
queue<int> q;
q.push(s);
while (!q.empty())
{
//从队列中拉取一个节点
s = q.front();
q.pop();

//看 s 自己是否向上级建立了通道
if (build[s])
{
ret++;
build[s] =false;
}
//遍历 s 所有的下级
for (int i = 0; i < down[s].size(); ++i)
{
// 看这个下级是否建立了通道
if (build[down[s][i]])
q.push(down[s][i]);
}
}
return ret;
}

int main()
{
int i;
cin >> n;
for (i = 1; i <n; ++i)
{
scanf("%d", &up[i]);     // 设置 i的上级
down[up[i]].push_back(i); // 设置属于up[i] 的下级
}
char buffer[20];
scanf("%d",&m);
for (i = 0; i <m; ++i)
{
int num;
scanf("%s", buffer);
scanf("%d", &num);
//只用比较第一个字符就可以知道是哪条命令
if (buffer[0] =='S')
printf("%d\n", Send(num));
else
printf("%d\n", Danger(num));
}
return 0;
}

C：最小秘钥

题目链接

正解：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int T,n,num[3009];
scanf("%d",&T);
int A[20009];
while(T--)
{
int n,flag=0;
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d",&num[i]);
for(int i=n;;i++)
{
flag=0;
memset(A,0,sizeof(A));
for(int j=0;j<n;j++)
{
int op=num[j]%i;
if(A[op]==0)
A[op]=1;
else
{
flag=1;
break;
}
}
if(flag==0)
{
printf("%d\n",i);
break;
}
}
}
return 0;
}

D：年终奖金

题目链接

正解：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1<<31-1
#define min(x,y)(x<y?x:y)
using namespace std;
int a[110];
int dp[110];
int get(int x,int y)
{
return (x-y)*(x-y);
}
int main()
{
int n,k,c,i,j;
while(scanf("%d%d%d",&n,&k,&c)!=EOF)
{
for(i=1;i<=n;i++)
{
dp[i]=MAX;
}
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+1+n);
dp[0]=0;
for(i=k;i<=n;i++)
{
for(j=0;j<=i-k;j++)
{
if(j<k&&j!=0)
continue;
dp[i]=min(dp[j]+c+get(a[i],a[j+1]),dp[i]);
}
}
printf("%d\n",dp
);
}
return 0;
}

E：GDP值

题目链接

正解：

#include <iostream>
#include <cstdio>
#include <set>
#include<algorithm>
#include<cstdlib>
#include<bitset>
#include<cstring>
using namespace std;
const int MAXN = 1024;
typedef bitset<MAXN>bst;
struct node
{
int x, next;
bst v;
} e[MAXN << 1];
char ch[MAXN];
int n, m, P, cnt, len, Link[MAXN], f[MAXN], wh[MAXN], E[MAXN][2];
bst mat[MAXN], M[MAXN], dis[MAXN], V[MAXN];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')
{
f = -1;
}
ch = getchar();
}
while (isdigit(ch))
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}

bst Read()
{
bst temp;
temp.reset();
scanf("%s", ch + 1);
int len = (int)strlen(ch + 1);
for (int i = 1; i <= len; i++)
{
temp[len - i] = ch[i] - '0';
}
return temp;
}

void insert(int x, int y, bst v)
{
e[++len].next = Link[x];
Link[x] = len;
e[len].x = y;
e[len].v = v;
e[++len].next = Link[y];
Link[y] = len;
e[len].x = x;
e[len].v = v;
}
int find(int x)
{
return f[x] == x ? x : f[x] = find(f[x]);
}
void dfs(int x, int pre)
{
for (int i = Link[x]; i; i = e[i].next)
{
if (e[i].x != pre)
{
dis[e[i].x] = dis[x] ^ e[i].v;
dfs(e[i].x, x);
}
}
}
void updata(int x, bst y)
{
int pos = 0;
for (int i = 1; i <= 1000; ++i)
{
if (M[i][x] && mat[i].none())
{
pos = i;
break;
}
}
if (!pos)
{
for (int i = 0; i <= 1000; ++i)
{
if (M[wh[i]][x] && wh[i] != 0)
{
pos = wh[i];
wh[i] = 0;
break;
}
}
}
for (int i = 1; i <= 1000; ++i)
{
if (pos != i && M[i][x])
{
mat[i] = mat[i] ^ mat[pos], M[i] = M[i] ^ M[pos];
}
}
mat[pos] ^= y;
for (int i = 1000; i >= 0; --i)
{
if (mat[pos][i])
{
if (!wh[i])
{
wh[i] = pos;
break;
}
else
{
mat[pos] = mat[pos] ^ mat[wh[i]];
M[pos] = M[pos] ^ M[wh[i]];
}
}
}
}
void output()
{
bst ans;
ans.reset();
char ts[MAXN], *tmp;
for (int i = 1000; i >= 0; --i)
{
if (ans[i] == 0 && wh[i])
{
ans = ans ^ mat[wh[i]];
}
}
if (ans.none())
{
puts("0");
}
else
{
tmp = ts;
for (int i = 1000; i >= 0; --i)
{
if (ans[i])
{
for (int j = i; j >= 0; --j)
{
*(tmp++) = ans[j] + '0';
}
break;
}
}
*(tmp++) = '\0';
puts(ts);
}
}
int main()
{
n = read();
m = read();
P = read();
for (int i = 1; i <= n; ++i)
{
f[i] = i;
}
for (int i = 1; i <= 1000; ++i)
{
M[i][i] = 1;
}
for (int i = 1; i <= m; ++i)
{
int x = read(), y = read();
bst value = Read();
if (find(x) == find(y))
{
E[++cnt][0] = x, E[cnt][1] = y, V[cnt] = value;
}
else
{
f[find(x)] = find(y), insert(x, y, value);
}
}
dfs(1, 0);
int CASE = cnt;
for (int i = 1; i <= cnt; ++i)
{
updata(i, dis[E[i][0]] ^ dis[E[i][1]] ^ V[i]);
}
output();
for (int i = 1; i <= P; ++i)
{
char opt[10];
scanf("%s", opt);
if (opt[1] == 'd')
{
int x = read(), y = read();
bst value = Read();
E[++cnt][0] = x;
E[cnt][1] = y;
V[cnt] = value;
updata(cnt, dis[E[cnt][0]] ^ dis[E[cnt][1]] ^ V[cnt]);
}
else if (opt[1] == 'h')
{
int x = read();
bst value = Read();
updata(x + CASE, V[x + CASE] ^ value);
V[x + CASE] = value;
}
else
{
int x = read();
updata(x + CASE, dis[E[x + CASE][0]] ^ dis[E[x + CASE][1]] ^ V[x + CASE]);
V[x + CASE] = dis[E[x + CASE][0]] ^ dis[E[x + CASE][1]];
}
output();
}
return 0;
}

F：Binary to Prime

题目链接

正解

#include<bits/stdc++.h>
using namespace std;
int main()
{
int a[160];
a[1]=2;
a[2]=3;
int k=3;
for(int i=4;;i++)
{
int t=1;
for(int j=2;j<=sqrt(i);j++)
{
if(i%j==0)
{
t=0;
break;
}
}
if(t==1)
{
a[k]=i;
k++;
}
if(k>155)
break;
}
char b[200];
while(scanf("%s",b)!=EOF)
{
int l=strlen(b);
int sum=0;
for(int i=l-1;i>=0;i--)
{

c144
if(b[i]=='1')
{
sum=sum+a[l-i];
}
}
printf("%d\n",sum);
}
return 0;
}

G:Plumbing the depth of lake

题目链接

正解：

#include <cstdio>
#include <cstring>
int map[55][55];
int dx[8] = {-1,-1,-1,0,0,1,1,1};
int dy[8] = {-1,0,1,-1,1,-1,0,1};
int max;
void search(int i,int j)
{
int k;
if(map[i][j]==-1) return ;

if(max >= map[i][j]) return ;

for(k=0; k<8; k++)
{
if(map[i][j] == map[i + dx[k]] [j + dy[k]]) //八个方向找一下,有就更新 max
{
max = map[i][j];
break;
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
max = -1;
int i,j,m,n;
scanf("%d%d",&m,&n);
memset(map,-1,sizeof(map));
for(i=1; i<=m; i++)
{
for(j=1; j<=n; j++)
{
scanf("%d",&map[i][j]);
}
}
for(i=1; i<=m; i++)
{
for(j=1; j<=n; j++)
{
search(i,j);
}
}
printf("%d\n",max);
}
return 0;
}

H:Intelligent Parking Building

题目链接

正解：

#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
struct Node
{
int floor;
int num;
} node[2509];
int main()
{
int T,n,m,Max;
scanf("%d",&T);
while(T--)
{
Max=-1;
memset(node,NULL,sizeof(node));
int Tu[52][52];
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)
{
Tu[i][0] = 1;
for(int j=1; j<=m; j++)
{
scanf("%d",&Tu[i][j]);
if(Tu[i][j]!=-1)
{
if(Tu[i][j]>Max)
Max=Tu[i][j];
node[Tu[i][j]].floor=i;
node[Tu[i][j]].num=j;
}
}
}
int  sum=0,flag;
for(int i=1; i<=Max; i++)
{
sum+=(node[i].floor-1)*10*2;
// cout<<"sum1==="<<sum<<endl;
int mm=min(node[i].num-1,1+m-node[i].num);
int res = min(abs(node[i].num - Tu[node[i].floor][0]),m- abs(node[i].num - Tu[node[i].floor][0]));   //在同行中 找距离左右两边最近的那个。
sum += res * 5;
Tu[node[i].floor][0]=node[i].num;
}
printf("%d\n",sum);
}
return 0;
}

I：Trainsmit information

题目链接：

正解：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<map>
#define inf 0x3f3f3f3f

using namespace std;

typedef long long ll;
const int maxn = 110;
int N,M,S,E;   ///经过N个间谍传播，M道路条数，
struct matrix
{
ll a[maxn][maxn];
}dis,ans;
matrix flody(matrix A,matrix B,int n)
{
matrix tmp;
for(int i = 1; i < n; i++)
for(int j = 1; j < n; j++)
tmp.a[i][j] = inf;
for(int k = 1; k < n; k++)
for(int i = 1; i < n; i++)
for(int j = 1; j < n; j++)
if(tmp.a[i][j]>A.a[i][k]+B.a[k][j])
tmp.a[i][j] = A.a[i][k] + B.a[k][j];
return tmp;
}
void solve(int n,int s,int e)
{
for(int i = 1; i < n; i++)
for(int j = 1; j < n; j++)
ans.a[i][j] = dis.a[i][j];
N--;  ///因为ans[i][j]是一步的距离，相当于已经传给一个间谍了
while(N)
{
if(N&1)
{
ans = flody(ans,dis,n);
}
dis = flody(dis,dis,n);
N = N>>1;
}
if(ans.a[s][e] < inf)
printf("%lld\n",ans.a[s][e]);
}
int main()
{
int T,len,x,y;
scanf("%d",&T);
while(T--)
{
map<int,int>m;
int num = 1;
scanf("%d%d%d%d",&N,&M,&S,&E);
for(int i = 0; i < maxn; i++)
for(int j = 0; j < maxn; j++)
dis.a[i][j] = inf;
for(int i = 1; i <= M; i++)
{
scanf("%d%d%d",&len,&x,&y);
if(m[x]==0)
{
m[x] = num++;
}
int xi = m[x];
if(m[y]==0)
{
m[y] = num++;
}
int yi = m[y];
if(len < dis.a[xi][yi])
dis.a[xi][yi] = dis.a[yi][xi] = len;
}
solve(num,m[S],m[E]);
}
return 0;
}