整理第十届河南省ACM省赛正解
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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[N].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[n]); } 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--) { 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;}
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- 整理第十届河南省ACM省赛正解
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- 河南省第十届ACM省赛题目：问题 A: 谍报分析
- 河南省第十届ACM省赛题目：问题 F: Binary to Prime
- 河南省第十届ACM省赛题目：问题 G: Plumbing the depth of lake
- 河南省第十届ACM省赛题目：问题 C: 最小秘钥
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