bzoj 一些题目汇总

2140: 稳定婚姻

 /*
 求联通分量。
 */
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 int head[N],nxt[],to[];
 int dfn[N],low[N],st[N],bel[N];
 int Enum,TimeIndex,NumberPeople,BlockIndex,top;
 bool vis[N];

 map<string,int> p;

 struct Couple{
     string a,b;
 }c[N];

 inline void add_edge(int u,int v) {
     ++Enum;to[Enum] = v, nxt[Enum] = head[u],head[u] = Enum;
 }

 void tarjan(int u) {
     dfn[u] = low[u] = ++TimeIndex;
     st[++top] = u; // --
     vis[u] = true; // --
     for (int i=head[u]; i; i=nxt[i]) {
         int v = to[i];
         if (!dfn[v]) {
             tarjan(v);
             low[u] = min(low[u],low[v]);
         }
         else if (vis[v]) low[u] = min(low[u],dfn[v]);
     }
     if (dfn[u] == low[u]) {
         ++BlockIndex;
         do {
             bel[st[top]] = BlockIndex;
             vis[st[top]] = false; // --
             top--;
         } while (st[top+] != u);
     }
 }

 int main() {
     int n = read();
     string a,b;
     int tot = ;
     for (int i=; i<=n; ++i) {
         cin >> c[i].a >> c[i].b;
         p[c[i].a] = ++NumberPeople;
         p[c[i].b] = ++NumberPeople;
         add_edge(NumberPeople-,NumberPeople);
     }
     int m = read();
     for (int i=; i<=m; ++i) {
         cin >> a >> b;
         add_edge(p[b],p[a]);
     }
     for (int i=; i<=NumberPeople; ++i) { // -- i<=n
         if (!dfn[i]) tarjan(i);
     }
     for (int i=; i<=n; ++i) {
         if (bel[p[c[i].a]] == bel[p[c[i].b]]) puts("Unsafe");
         else puts("Safe");
     }
     return ;
 }

2783: [JLOI2012]树

 /*
 读好题目。
 所有路径都是 深度小的->深度大的。
 所以，dfs一遍。
 */
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 int head[N],nxt[N<<],to[N<<],Enum;
 int w[N],S,Ans;
 map<int,int> cnt;

 inline void add_edge(int u,int v) {
     ++Enum;to[Enum] = v, nxt[Enum] = head[u],head[u] = Enum;
 }

 void dfs(int u,int fa,int sum) {
     if (cnt[sum - S]) Ans += cnt[sum - S];
     for (int i=head[u]; i; i=nxt[i]) {
         int v = to[i];
         if (v == fa) continue;
         cnt[sum+w[v]] ++;
         dfs(v,u,sum+w[v]);
         cnt[sum+w[v]] --;
     }
 }

 int main() {
     int n = read();S = read();
     for (int i=; i<=n; ++i) w[i] = read();
     for (int i=; i<n; ++i) {
         int u = read(),v = read();
         add_edge(u,v);add_edge(v,u);
     }
     cnt[] = ;cnt[w[]] = ;
     dfs(,,w[]);
     cout << Ans;
     return ;
 }

2429: [HAOI2006]聪明的猴子

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 struct Edge{
     int u,v;double w;
     Edge() {}
     Edge(int a,int b,double c) {u = a, v = b, w = c;} // double c
     bool operator < (const Edge &A) const {
         return w < A.w;
     }
 }e[N*N];
 int x[N],y[N],fa[N],a[N],Enum;

 int find(int x) {
     if (x == fa[x]) return x;
     return fa[x] = find(fa[x]);
 }
 int main() {
     int n = read();
     for (int i=; i<=n; ++i) a[i] = read();
     sort(a+,a+n+);
     int m = read();
     for (int i=; i<=m; ++i)
         x[i] = read(),y[i] = read();
     for (int i=; i<=m; ++i)
         for (int j=i+; j<=m; ++j) {
             double w = sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j]));
             e[++Enum] = Edge(i,j,w);
         }
     sort(e+,e+Enum+);
     for (int i=; i<=m; ++i) fa[i] = i; //-- i<=m
     int cnt = ;double mx;
     for (int i=; i<=Enum; ++i) {
         int u = find(e[i].u), v = find(e[i].v);
         if (u != v) {
             fa[u] = v;
             cnt ++;
             mx = e[i].w;
             if (cnt == m - ) break; // -- cnt=n-1
         }
     }
     int Ans = ;
     for (int i=n; i>=; --i) if (a[i] >= mx) Ans ++; //--居然反了。。
     cout << Ans;
     return ;
 }

2946: [Poi2000]公共串

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;

 struct SuffixAutomaton{
     int Last, Index, res, cur, fa[N], trans[N][], len[N];
     SuffixAutomaton() {Last = Index = cur = ; res = ;}
     void extend(int c) {
         int P = Last, NP = ++Index;
         len[NP] = len[P] + ;
         for (; P&&!trans[P][c]; P=fa[P]) trans[P][c] = NP;
         if (!P) fa[NP] = ;
         else {
             int Q = trans[P][c];
             if (len[P] +  == len[Q]) fa[NP] = Q;
             else {
                 int NQ = ++Index;
                 fa[NQ] = fa[Q];
                 len[NQ] = len[P] + ;
                 memcpy(trans[NQ], trans[Q], sizeof trans[Q]);
                 fa[Q] = NQ;
                 fa[NP] = NQ;
                 for (; P&&trans[P][c]==Q; P=fa[P]) trans[P][c] = NQ;
             }
         }
         Last = NP;
     }
     int solve(int c) {
         if (trans[cur][c]) {cur = trans[cur][c]; res++; return res;}
         for (; cur&&!trans[cur][c]; cur=fa[cur]);
         if (!cur) res = , cur = ;
         else res = len[cur] + , cur = trans[cur][c];
         return res;
     }
 }sam[];

 char s[N];
 char str[N];

 int main() {
     int n = ,t = ,len;
     scanf("%s",str+);

     while (scanf("%s",s+)!=EOF) {
         len = strlen(s + );
         for (int i=; i<=len; ++i)
             sam[t].extend(s[i] - 'a');
         t ++;
     }
     int ans = ;
     len = strlen(str+);
     for (int i=; i<=len; ++i) {
         int tmp = 1e9;
         for (int j=; j<t; ++j)
             tmp = min(tmp, sam[j].solve(str[i] - 'a'));
         ans = max(ans, tmp);
     }
     printf("%d",ans);
     return ;
 }

3613: [Heoi2014]南园满地堆轻絮

 /*
 二分，每个数对应一个区间，每次取区间的最小值。
 */
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 LL A[N], Sa, Sb, Sc, Sd, mod;
 int n;

 LL F(LL x) {
     LL x2 = x * x % mod, x3 = x2 * x % mod; //---
     return (Sa * x3 % mod + Sb * x2 % mod + Sc * x % mod + Sd) % mod;
 }
 void init() {
     Sa = read(), Sb = read(), Sc = read(), Sd = read(); A[] = read(); mod = read();
     for (int i=; i<=n; ++i)
         A[i] = (F(A[i-]) + F(A[i-])) % mod;
 }

 bool check(LL x) {
     LL last = A[] - x;
     for (int i=; i<=n; ++i) {
         if (A[i] >= last) {
             last = max(A[i] - x, last);
         }
         else {
             if (A[i] + x < last) return false; //---
             last = min(A[i] + x, last);
         }
     }
     return true;
 }

 int main() {

     n = read();
     init();

     LL L = , R = mod, ans; //---
     while (L <= R) {
         LL mid = (L + R) >> ;
         if (check(mid)) ans = mid, R = mid - ;
         else L = mid + ;
     }
     cout << ans;

     return ;
 }

 /*
 找出最大的逆序对，然后答案是(mx-mn+1)/2
 如果将最大的逆序对可以提升的一个平台，那么其他的逆序对也都可在这一平台。

 下面摘自https://blog.csdn.net/vmurder/article/details/44096565
 我们把所有逆序对点都搞到同一高度。
 然后发现答案是距离最远的逆序对搞到一起的代价。
 */
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for (;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 LL A[N], Sa, Sb, Sc, Sd, mod;
 int n;

 LL F(LL x) {
     LL x2 = x * x % mod, x3 = x2 * x % mod;
     return (Sa * x3 % mod + Sb * x2 % mod + Sc * x % mod + Sd) % mod;
 }
 void init() {
     Sa = read(), Sb = read(), Sc = read(), Sd = read(); A[] = read(); mod = read();
     for (int i=; i<=n; ++i)
         A[i] = (F(A[i-]) + F(A[i-])) % mod;
 }

 int main() {

     n = read();
     init();

     LL Mx = -1e9, ans = ;
     for (int i=; i<=n; ++i) {
         if (A[i] >= Mx) Mx = A[i];
         else ans = max(ans, (Mx - A[i] + ) / );
     }   

     cout << ans;

     return ;
 }

-----------