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  • bzoj 一些题目汇总

    2140: 稳定婚姻

     1 /*
     2 求联通分量。
     3 */
     4 #include<bits/stdc++.h>
     5 using namespace std;
     6 typedef long long LL;
     7  
     8 inline int read() {
     9     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    10     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
    11 }
    12  
    13 const int N = 10010;
    14 int head[N],nxt[50010],to[50010];
    15 int dfn[N],low[N],st[N],bel[N];
    16 int Enum,TimeIndex,NumberPeople,BlockIndex,top;
    17 bool vis[N];
    18  
    19  
    20 map<string,int> p;
    21  
    22 struct Couple{
    23     string a,b;
    24 }c[N];
    25  
    26 inline void add_edge(int u,int v) {
    27     ++Enum;to[Enum] = v, nxt[Enum] = head[u],head[u] = Enum;
    28 }
    29  
    30 void tarjan(int u) {
    31     dfn[u] = low[u] = ++TimeIndex;
    32     st[++top] = u; // --
    33     vis[u] = true; // --
    34     for (int i=head[u]; i; i=nxt[i]) {
    35         int v = to[i];
    36         if (!dfn[v]) {
    37             tarjan(v);
    38             low[u] = min(low[u],low[v]);
    39         }
    40         else if (vis[v]) low[u] = min(low[u],dfn[v]);
    41     }
    42     if (dfn[u] == low[u]) {
    43         ++BlockIndex;
    44         do {
    45             bel[st[top]] = BlockIndex;
    46             vis[st[top]] = false; // --
    47             top--;
    48         } while (st[top+1] != u);
    49     }
    50 }
    51  
    52 int main() {
    53     int n = read();
    54     string a,b;
    55     int tot = 0;
    56     for (int i=1; i<=n; ++i) {
    57         cin >> c[i].a >> c[i].b;
    58         p[c[i].a] = ++NumberPeople;
    59         p[c[i].b] = ++NumberPeople;
    60         add_edge(NumberPeople-1,NumberPeople);
    61     }
    62     int m = read();
    63     for (int i=1; i<=m; ++i) {
    64         cin >> a >> b;
    65         add_edge(p[b],p[a]);
    66     }
    67     for (int i=1; i<=NumberPeople; ++i) { // -- i<=n
    68         if (!dfn[i]) tarjan(i);
    69     }
    70     for (int i=1; i<=n; ++i) {
    71         if (bel[p[c[i].a]] == bel[p[c[i].b]]) puts("Unsafe");
    72         else puts("Safe");
    73     }
    74     return 0;
    75 }
    View Code

    2783: [JLOI2012]树

     1 /*
     2 读好题目。
     3 所有路径都是 深度小的->深度大的。 
     4 所以,dfs一遍。
     5 */
     6 #include<bits/stdc++.h>
     7 using namespace std;
     8 typedef long long LL;
     9   
    10 inline int read() {
    11     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    12     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
    13 }
    14  
    15 const int N = 100005;
    16 int head[N],nxt[N<<1],to[N<<1],Enum;
    17 int w[N],S,Ans;
    18 map<int,int> cnt;
    19  
    20 inline void add_edge(int u,int v) {
    21     ++Enum;to[Enum] = v, nxt[Enum] = head[u],head[u] = Enum;
    22 }
    23  
    24 void dfs(int u,int fa,int sum) {
    25     if (cnt[sum - S]) Ans += cnt[sum - S];
    26     for (int i=head[u]; i; i=nxt[i]) {
    27         int v = to[i];
    28         if (v == fa) continue;
    29         cnt[sum+w[v]] ++;
    30         dfs(v,u,sum+w[v]);
    31         cnt[sum+w[v]] --;
    32     }
    33 }
    34  
    35 int main() {
    36     int n = read();S = read();
    37     for (int i=1; i<=n; ++i) w[i] = read();
    38     for (int i=1; i<n; ++i) {
    39         int u = read(),v = read();
    40         add_edge(u,v);add_edge(v,u);
    41     }
    42     cnt[0] = 1;cnt[w[1]] = 1;
    43     dfs(1,0,w[1]);
    44     cout << Ans;
    45     return 0;
    46 }
    View Code

    2429: [HAOI2006]聪明的猴子

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4  
     5 inline int read() {
     6     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
     7     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
     8 }
     9  
    10 const int N = 1010;
    11 struct Edge{
    12     int u,v;double w;
    13     Edge() {}
    14     Edge(int a,int b,double c) {u = a, v = b, w = c;} // double c
    15     bool operator < (const Edge &A) const {
    16         return w < A.w;
    17     }
    18 }e[N*N];
    19 int x[N],y[N],fa[N],a[N],Enum;
    20  
    21 int find(int x) {
    22     if (x == fa[x]) return x;
    23     return fa[x] = find(fa[x]);
    24 }
    25 int main() {
    26     int n = read();
    27     for (int i=1; i<=n; ++i) a[i] = read();
    28     sort(a+1,a+n+1);
    29     int m = read();
    30     for (int i=1; i<=m; ++i) 
    31         x[i] = read(),y[i] = read();
    32     for (int i=1; i<=m; ++i) 
    33         for (int j=i+1; j<=m; ++j) {
    34             double w = sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j]));
    35             e[++Enum] = Edge(i,j,w);
    36         }
    37     sort(e+1,e+Enum+1);
    38     for (int i=1; i<=m; ++i) fa[i] = i; //-- i<=m
    39     int cnt = 0;double mx;
    40     for (int i=1; i<=Enum; ++i) { 
    41         int u = find(e[i].u), v = find(e[i].v);
    42         if (u != v) { 
    43             fa[u] = v;
    44             cnt ++;
    45             mx = e[i].w;
    46             if (cnt == m - 1) break; // -- cnt=n-1
    47         }
    48     }
    49     int Ans = 0;
    50     for (int i=n; i>=1; --i) if (a[i] >= mx) Ans ++; //--居然反了。。 
    51     cout << Ans;
    52     return 0;
    53 }
    View Code

    2946: [Poi2000]公共串

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 
     5 inline int read() {
     6     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
     7     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
     8 }
     9 
    10 const int N = 200010;
    11 
    12 struct SuffixAutomaton{
    13     int Last, Index, res, cur, fa[N], trans[N][26], len[N];
    14     SuffixAutomaton() {Last = Index = cur = 1; res = 0;}
    15     void extend(int c) {
    16         int P = Last, NP = ++Index;
    17         len[NP] = len[P] + 1;
    18         for (; P&&!trans[P][c]; P=fa[P]) trans[P][c] = NP;
    19         if (!P) fa[NP] = 1;
    20         else {
    21             int Q = trans[P][c];
    22             if (len[P] + 1 == len[Q]) fa[NP] = Q;
    23             else {
    24                 int NQ = ++Index;
    25                 fa[NQ] = fa[Q];
    26                 len[NQ] = len[P] + 1;
    27                 memcpy(trans[NQ], trans[Q], sizeof trans[Q]);
    28                 fa[Q] = NQ;
    29                 fa[NP] = NQ;
    30                 for (; P&&trans[P][c]==Q; P=fa[P]) trans[P][c] = NQ;
    31             }
    32         }
    33         Last = NP;
    34     }
    35     int solve(int c) {
    36         if (trans[cur][c]) {cur = trans[cur][c]; res++; return res;}
    37         for (; cur&&!trans[cur][c]; cur=fa[cur]);
    38         if (!cur) res = 0, cur = 1;
    39         else res = len[cur] + 1, cur = trans[cur][c];
    40         return res;
    41     }
    42 }sam[9];
    43 
    44 char s[N];
    45 char str[N];
    46 
    47 int main() {
    48     int n = 0,t = 0,len;
    49     scanf("%s",str+1);
    50     
    51     while (scanf("%s",s+1)!=EOF) {
    52         len = strlen(s + 1);
    53         for (int i=1; i<=len; ++i) 
    54             sam[t].extend(s[i] - 'a');
    55         t ++;
    56     }
    57     int ans = 0;
    58     len = strlen(str+1);
    59     for (int i=1; i<=len; ++i) {
    60         int tmp = 1e9;
    61         for (int j=0; j<t; ++j) 
    62             tmp = min(tmp, sam[j].solve(str[i] - 'a'));
    63         ans = max(ans, tmp);
    64     }
    65     printf("%d",ans);
    66     return 0;
    67 }
    View Code

    3613: [Heoi2014]南园满地堆轻絮

     1 /*
     2 二分,每个数对应一个区间,每次取区间的最小值。
     3 */
     4 #include<bits/stdc++.h>
     5 using namespace std;
     6 typedef long long LL;
     7 
     8 inline int read() {
     9     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    10     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
    11 }
    12 
    13 const int N = 5000100;
    14 LL A[N], Sa, Sb, Sc, Sd, mod;
    15 int n;
    16 
    17 LL F(LL x) {
    18     LL x2 = x * x % mod, x3 = x2 * x % mod; //---
    19     return (Sa * x3 % mod + Sb * x2 % mod + Sc * x % mod + Sd) % mod;
    20 }
    21 void init() {
    22     Sa = read(), Sb = read(), Sc = read(), Sd = read(); A[1] = read(); mod = read();
    23     for (int i=2; i<=n; ++i) 
    24         A[i] = (F(A[i-1]) + F(A[i-2])) % mod;
    25 }
    26 
    27 bool check(LL x) {
    28     LL last = A[1] - x;
    29     for (int i=2; i<=n; ++i) {
    30         if (A[i] >= last) {
    31             last = max(A[i] - x, last);
    32         }
    33         else {
    34             if (A[i] + x < last) return false; //---
    35             last = min(A[i] + x, last);
    36         }
    37     }
    38     return true;
    39 }
    40 
    41 int main() {
    42     
    43     n = read();
    44     init();
    45     
    46     LL L = 0, R = mod, ans; //---
    47     while (L <= R) {
    48         LL mid = (L + R) >> 1;
    49         if (check(mid)) ans = mid, R = mid - 1;
    50         else L = mid + 1;
    51     }
    52     cout << ans;
    53     
    54     return 0;
    55 }
    View Code
     1 /*
     2 找出最大的逆序对,然后答案是(mx-mn+1)/2
     3 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。
     4 
     5 下面摘自https://blog.csdn.net/vmurder/article/details/44096565 
     6 我们把所有逆序对点都搞到同一高度。 
     7 然后发现答案是距离最远的逆序对搞到一起的代价。
     8 */
     9 #include<bits/stdc++.h>
    10 using namespace std;
    11 typedef long long LL;
    12  
    13 inline int read() {
    14     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    15     for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
    16 }
    17  
    18 const int N = 5000100;
    19 LL A[N], Sa, Sb, Sc, Sd, mod;
    20 int n;
    21  
    22 LL F(LL x) {
    23     LL x2 = x * x % mod, x3 = x2 * x % mod;
    24     return (Sa * x3 % mod + Sb * x2 % mod + Sc * x % mod + Sd) % mod;
    25 }
    26 void init() {
    27     Sa = read(), Sb = read(), Sc = read(), Sd = read(); A[1] = read(); mod = read();
    28     for (int i=2; i<=n; ++i) 
    29         A[i] = (F(A[i-1]) + F(A[i-2])) % mod;
    30 }
    31  
    32 int main() {
    33      
    34     n = read();
    35     init();
    36      
    37     LL Mx = -1e9, ans = 0;
    38     for (int i=1; i<=n; ++i) {
    39         if (A[i] >= Mx) Mx = A[i];
    40         else ans = max(ans, (Mx - A[i] + 1) / 2);
    41     }   
    42 
    43     cout << ans;
    44      
    45     return 0;
    46 }
    View Code

    -----------

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  • 原文地址:https://www.cnblogs.com/mjtcn/p/9281180.html
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