2010 福州区域赛 练习赛
2015-05-17 23:18:26
总结:这场开局不太顺利... 还好后程发力,过掉了一些中等题。
赛中5题,赛后目前6题... 正在补ing...
先写这两道题吧。最近有点忙... 要先练练技巧 & 手速题了。
B题 hdu 3691: 全局最小割,Stoer_Wagner
这题... 在比赛中可谓苦思冥想,首先枚举汇点跑最大流的话,就算用上最快的最高标号先流推进算法也需要 O(n^3 * sqrt(m) ),更不用提
用 Dinic 跑的 O(n^3 * m) 了。想过跟树肯定有关系,但是由于并没有写过这样的题 ... GG (以上小吐槽... 忽略)
思路:正解其实就是Stoer_Wagner算法,裸的全局最小割,注意到答案和起点并没有关系,所以直接跑一边就出答案了 TAT。
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; const int MAXN = 310; ll g[MAXN][MAXN]; ll wg[MAXN]; int v[MAXN]; bool vis[MAXN]; int N,M,K; int SW(int n){ for(int i = 1; i <= n; ++i) v[i] = i; ll res = -1; while(n > 1){ memset(vis,false,sizeof(vis)); memset(wg,0,sizeof(wg)); int pre = 1; vis[pre] = true; for(int i = 1; i < n; ++i){ int p = -1; for(int j = 2; j <= n; ++j) if(!vis[v[j]]){ wg[v[j]] += g[v[pre]][v[j]]; if(p == -1 || wg[v[j]] > wg[v[p]]) p = j; } vis[v[p]] = true; //printf("n : %d, %d %d\n",n,v[pre],v[p]); if(i == n - 1){ if(res == -1) res = wg[v[p]]; else res = min(res,wg[v[p]]); //printf("n : %d , res : %d\n",n,res); for(int j = 1; j <= N; ++j){ g[v[pre]][v[j]] += g[v[p]][v[j]]; g[v[j]][v[pre]] += g[v[j]][v[p]]; } v[p] = v[n--]; } pre = p; } } return res; } int main(){ int a,b,c; while(scanf("%d%d%d",&N,&M,&K) != EOF){ if(!N && !M && !K) break; memset(g,0,sizeof(g)); for(int i = 1; i <= M; ++i){ scanf("%d%d%d",&a,&b,&c); g[a][b] += c; g[b][a] += c; } printf("%lld\n",SW(N)); } return 0; }
F题 hdu 3695:AC 自动机
题意:给出 n 个模式串(串的长度不超过1000),再以压缩的形式给出文本串(解压缩后最长510w),模式串倒过来在文本串中匹配也算匹配。
思路:将文本串暴力解压缩(但是注意 [ ] 中的数字 > 1000 的话,就把它设为1000,因为模式串最长才1000)然后跑一遍匹配,再把文本串倒序后跑一遍匹配即可。
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <map> #include <set> #include <stack> #include <queue> #include <string> #include <iostream> #include <algorithm> using namespace std; #define getmid(l,r) ((l) + ((r) - (l)) / 2) #define MP(a,b) make_pair(a,b) #define PB(a) push_back(a) typedef long long ll; typedef pair<int,int> pii; const double eps = 1e-8; const int INF = (1 << 30) - 1; const int MAXN = 250010; const int RA = 26; int T,N,vis[260]; char tmp[1010],S[5100010],s[5100010]; struct AC_auto{ int next[MAXN][RA],f[MAXN],cnt[MAXN],last[MAXN]; int root,tot; void new_node(int p){ for(int i = 0; i < RA; ++i) next[p][i] = 0; f[p] = cnt[p] = last[p] = 0; } void init(){ root = tot = 0; new_node(root); } void insert(int k,char *str){ int len = strlen(str); int p = root; for(int i = 0; i < len; ++i){ int id = str[i] - 'A'; if(!next[p][id]){ next[p][id] = ++tot; new_node(tot); } p = next[p][id]; } cnt[p] = k; } void getfail(){ queue<int> Q; f[root] = root; for(int i = 0; i < RA; ++i){ int u = next[root][i]; if(u) Q.push(u); } while(!Q.empty()){ int x = Q.front(); Q.pop(); for(int i = 0; i < RA; ++i){ int u = next[x][i]; if(!u){ next[x][i] = next[f[x]][i]; continue; } Q.push(u); int v = f[x]; while(v && !next[v][i]) v = f[v]; f[u] = next[v][i]; last[u] = cnt[f[u]] ? f[u] : last[f[u]]; } } } void find(char *str){ int len = strlen(str); int p = root; for(int i = 0; i < len; ++i){ int id = str[i] - 'A'; p = next[p][id]; int cur = p; while(cur != root){ vis[cnt[cur]] = 1; cur = last[cur]; } } } }ac; int main(){ scanf("%d",&T); while(T--){ scanf("%d",&N); memset(vis,0,sizeof(vis)); ac.init(); for(int i = 1; i <= N; ++i){ scanf("%s",tmp); ac.insert(i,tmp); } ac.getfail(); scanf("%s",s); int len = strlen(s); int pos = 0; for(int i = 0; i < len; ++i){ if(s[i] == '['){ int val = 0; for(++i; s[i] >= '0' && s[i] <= '9'; ++i) val = val * 10 + s[i] - '0'; if(val > 1000) val = 1000; char c = s[i]; for(int j = 0; j < val; ++j) S[pos++] = c; ++i; } else S[pos++] = s[i]; } S[pos] = '\0'; ac.find(S); int top = pos / 2; char tc; for(int i = 0; i < top; ++i){ tc = S[i]; S[i] = S[pos - i - 1]; S[pos - i - 1] = tc; } S[pos] = '\0'; ac.find(S); int ans = 0; for(int i = 1; i <= N; ++i) if(vis[i]) ans++; printf("%d\n",ans); } return 0; }