有待搞懂
A. BNU 34067 Pair
题目链接： http://www.bnuoj.com/bnuoj/problem_show.php?pid=34067
#include <iostream> #include <cstdio> using namespace std; int main() { int n,t; scanf("%d",&t); while(t--) { scanf("%d",&n); int res = (2*n-1)/5; cout<<res<<endl; } return 0; }
B. Codeforces 371D Vessels
题目链接： http://acm.hust.edu.cn/vjudge/contest/view.action?cid=38994#problem/H
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <string> #include <vector> #include <map> #include <set> #include <time.h> #include <queue> #include <cctype> #include <utility> #include <numeric> #include <cstdlib> #include <iomanip> #include <sstream> #define Mod 1000000007 #define INT 2147483647 #define pi acos(-1.0) #define eps 1e-3 #define lll __int64 #define ll long long using namespace std; #define N 200010 int cap[N],jp[N],a[N]; int main() { int n,m; int i,j; int op,pos,val; int st; while(scanf("%d",&n)!=EOF) { memset(cap,0,sizeof(cap)); for(i=1;i<=n;i++) { scanf("%d",&a[i]); jp[i] = i; } scanf("%d",&m); for(i=0;i<m;i++) { scanf("%d",&op); if(op == 1) { scanf("%d%d",&pos,&val); st = jp[pos]; while(cap[st]+val>=a[st]&&st<=n) { val -= (a[st]-cap[st]); cap[st] = a[st]; st++; } if(st<=n) cap[st] += val; for(j=jp[pos];j<st;j++) { jp[j] = st; } jp[pos] = st; } else { scanf("%d",&pos); printf("%d\n",cap[pos]); } } } return 0; }
C. ZOJ 1518 The Sentence is False
题目：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=39061#problem/F
i&j表示第i个句子的内容是: Sentence j is true. i^j表示第i个句子的内容是: Sentence j is false.
当i&j时 i与j等价，当i^j时 i与j正好相反。利用并查集
若为: i&j 分3中情况: 1. i与j属于同一等价集合下一个循环 2.i与j的对立集合，或j与i的对立集合在同一个等价集合中.输出Inconsistent 3.合并i与j 合并 !i 与!j
若为i^j
合并i 与 !j 合并j 与 !i
最后循环每一个集合，对于每一个集合 取其 和 其对立集合的优集（对的元素多的集合）
代码：
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <utility> #include <cstdlib> using namespace std; #define N 1100 int fa[N],opp[N],num[N],vis[N]; void makeset(int n) { for(int i=1;i<=n;i++) { fa[i] = i; opp[i] = 0; num[i] = 1; } } int findset(int x) { if(x != fa[x]) { fa[x] = findset(fa[x]); } return fa[x]; } int main() { int i,j,n,flag,fai,faj,opi,opj,maxi; while(scanf("%d",&n)!=EOF && n) { makeset(n); flag = 0; for(i=1;i<=n;i++) { char s1[10],s2[10],s3[10]; scanf("%s%d%s%s",s1,&j,s2,s3); fai = findset(i); faj = findset(j); if(opp[fai]) //如果i的根有对手，找出对手的根 opi = findset(opp[fai]); else opi = 0; if(opp[faj]) //如果j的根有对手，找出对手的根 opj = findset(opp[faj]); else opj = 0; if(s3[0] == 't') //i说j是true { if(fai == faj) //如果两个在一个集合，成立 continue; if(opi == faj || fai == opj) //如果i,j相对，不符合 { flag = 1; //出现悖论，Inconsistent continue; } fa[faj] = fai; //否则合并梁集合 num[fai] += num[faj]; //权值相加 if(opi && opj) //如果对手都有 { fa[opj] = opi; //合并对手 num[opi] += num[opj]; } else if(opi) //如果只有i有对手 ** opp[faj] = opi; //j的根的对手赋为i的对手 ** else if(opj) //如果只有j有对手 ** opp[fai] = opj; //i的根的对手赋为j的对手 ** } else //i说j是错的 { if(opi == faj || opj == fai) //如果本来i和j是相对的，正确，继续 continue; if(fai == faj) //如果i和j是一个集合 { flag = 1; //悖论 continue; } if(opi) //如果i有对手，则i的对手和j是一个集合，合并 { fa[opi] = faj; num[faj] += num[opi]; } else //i没有对手 { opp[fai] = faj; //i集合的对手赋为j集合 } if(opj) //如果j有对手，则j的对手和i是一个集合，合并 { fa[opj] = fai; num[fai] += num[opj]; } else //j没有对手，则j集合的对手赋为i集合 { opp[faj] = fai; } } } if(flag) { printf("Inconsistent\n"); continue; } maxi = 0; memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) { if(i == fa[i] && !vis[i]) { if(opp[i]) { opi = findset(opp[i]); maxi += max(num[i],num[opi]); //取每个集合的本集合最大与其对立集合最大，求和 vis[opi] = 1; } else maxi += num[i]; vis[i] = 1; } } printf("%d\n",maxi); } return 0; }
D.POJ 1733 Parity game
题目：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=39052#problem/B
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <map> using namespace std; #define N 10100 int fa[N],num[N]; map<int,int> mp; void makeset(int n) { for(int i=1;i<=10000;i++) { fa[i] = i; num[i] = 0; } } int findset(int x) { if(x != fa[x]) { int tmp = fa[x]; fa[x] = findset(fa[x]); num[x] = num[x]^num[tmp]; } return fa[x]; } int unionset(int a,int b,char oe) { int x = findset(a); int y = findset(b); int k = (oe == 'o'? 1:0); if(x == y) { if(num[a]^num[b] == k) return 1; else return 0; } else { fa[x] = y; num[x] = num[a]^num[b]^k; return 1; } } int main() { int n,m,i,k,a,b,cnt; char ss[8]; scanf("%d%d",&n,&m); makeset(n); k = 1; cnt = 0; for(i=1;i<=m;i++) { scanf("%d%d %s",&a,&b,ss); a = a-1; if(mp.find(a) == mp.end()) { mp[a] = k++; } if(mp.find(b) == mp.end()) { mp[b] = k++; } if(unionset(mp[a],mp[b],ss[0])) cnt++; else break; } printf("%d\n",cnt); return 0; }
E.HDU 1867 A+B for you again
题目：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40277#problem/K
题意：
就是求str1 的最长后缀与 str2 的最长前缀。使得 str1+str2 的长度最小，并且字典序最小。
分析：
利用kmp 求出最长相同的后缀和前缀。在利用串比较函数比较最小字典序。
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 100007 char a[N],b[N]; int next1[N],next2[N]; void getnext(char *ss,int *next) { int i = 0,j = -1; next[0] = -1; int len = strlen(ss); while(i<len) { if(j == -1 || ss[i] == ss[j]) next[++i] = ++j; else j = next[j]; } } int kmp(char *s1,char *s2,int *next) { int i = 0,j = 0; int n = strlen(s1); int m = strlen(s2); while(i<n) // 为什么不加上 j<m ? { if(j == -1 || s1[i] == s2[j]) i++,j++; else j = next[j]; } return j; } int main() { int ka,kb; while(scanf("%s%s",a,b)!=EOF) { getnext(a,next1); getnext(b,next2); ka = kmp(a,b,next2); kb = kmp(b,a,next1); if(ka == kb) { if(strcmp(a,b)<=0) printf("%s%s\n",a,b+ka); else printf("%s%s\n",b,a+ka); } else if(ka>kb) printf("%s%s\n",a,b+ka); else printf("%s%s\n",b,a+kb); } return 0; }
F.2014 Topcoder Algorithm Round 1C --950
概率期望的数学解法。
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 57 class RedPaint { private: public: double expectedCells(int N) { double res = 1.0,cur = 1.0; int i; for(i=1;i<=N;i++) { if(i%2 == 0) cur = cur * (i-1)/i; res += cur; } return res; } };
G.HDU 4358 Boring counting
题解： http://www.cnblogs.com/wangfang20/archive/2013/05/24/3096620.html
四个modify
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <time.h> #include <cctype> #include <utility> #include <numeric> #include <functional> #include <iomanip> #include <sstream> #define Mod 1000000007 #define SMod m #define INint 2147483647 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-8 #define lll __int64 #define ll long long using namespace std; #define N 100007 struct node { int v,next; }G[2*N]; struct Query { int ind,l,r; }Q[N]; int head[2*N],tot,L[N],R[N],weight[N],Time,val[N],c[N],b[N],n,ans[N]; map<int,int> mp; vector<int> pre[N]; int lowbit(int x) { return x&-x; } int cmp(Query ka,Query kb) { return ka.r < kb.r; } void modify(int x,int val) { while(x <= n) { c[x] += val; x += lowbit(x); } } int getsum(int x) { int res = 0; while(x > 0) { res += c[x]; x -= lowbit(x); } return res; } void addedge(int u,int v) { G[tot].v = v; G[tot].next = head[u]; head[u] = tot++; } void dfs(int u) { L[u] = ++Time; val[u] = weight[u]; for(int i=head[u];i!=-1;i=G[i].next) { int v = G[i].v; if(!L[v]) dfs(v); } R[u] = Time; } int main() { int t,k,i,j,cs = 1,u,v,q; scanf("%d",&t); while(t--) { mp.clear(); scanf("%d%d",&n,&k); for(i=1;i<=n;i++) scanf("%d",&weight[i]),b[i] = weight[i]; sort(b+1,b+n+1); int ind = unique(b+1,b+n+1)-b; for(i=1;i<=ind;i++) mp[b[i]] = i; for(i=1;i<=n;i++) weight[i] = mp[weight[i]]; for(i=0;i<=ind;i++) pre[i].clear(),pre[i].push_back(0); memset(head,-1,sizeof(head)); memset(c,0,sizeof(c)); tot = Time = 0; memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); for(i=1;i<n;i++) { scanf("%d%d",&u,&v); addedge(u,v); addedge(v,u); } dfs(1); scanf("%d",&q); for(i=1;i<=q;i++) { scanf("%d",&u); Q[i].ind = i, Q[i].l = L[u], Q[i].r = R[u]; } sort(Q+1,Q+q+1,cmp); j = 1; for(i=1;i<=n;i++) { int v = val[i]; pre[v].push_back(i); int now = pre[v].size()-1; if(now >= k) { if(now > k) { modify(pre[v][now-k-1]+1,-1); modify(pre[v][now-k]+1,1); } modify(pre[v][now-k]+1,1); modify(pre[v][now-k+1]+1,-1); } while(Q[j].r == i) { ans[Q[j].ind] = getsum(Q[j].l); j++; } } printf("Case #%d:\n",cs++); for(i=1;i<=q;i++) printf("%d\n",ans[i]); if(t) puts(""); } return 0; }
H. Codeforces 396C On Changing Tree
题解： http://blog.csdn.net/night_raven/article/details/20239019
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #include <string> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <time.h> #include <cctype> #include <utility> #include <numeric> #include <functional> #include <iomanip> #include <sstream> #define Mod 1000000007 #define SMod m #define INint 2147483647 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-8 #define lll __int64 #define ll long long using namespace std; #define N 600017 int L[N],R[N],Time,n,m,layer,d[N]; lll c[2][N]; int head[N],tot; struct node { int v,next; }G[N]; void addedge(int u,int v) { G[tot].v = v; G[tot].next = head[u]; head[u] = tot++; } int lowbit(int x) { return x&-x; } void modify(int num,int x,lll val) { while(x <= 2*n) { c[num][x] = (c[num][x] + val)%Mod; x += lowbit(x); } } lll getsum(int num,int x) { lll res = 0; while(x > 0) { res = (res + c[num][x])%Mod; x -= lowbit(x); } return res; } void dfs(int u,int dep) { L[u] = ++Time; d[u] = dep; for(int i=head[u];i!=-1;i=G[i].next) { int v = G[i].v; if(!L[v]) dfs(v,dep+1); } R[u] = ++Time; } int main() { int i,j,u,v,op; lll x,k; while(scanf("%d",&n)!=EOF) { memset(c,0,sizeof(c)); memset(L,0,sizeof(L)); memset(R,0,sizeof(R)); memset(head,-1,sizeof(head)); tot = Time = 0; for(i=2;i<=n;i++) { scanf("%d",&u); addedge(u,i); addedge(i,u); } scanf("%d",&m); dfs(1,0); while(m--) { scanf("%d%d",&op,&u); if(op == 1) { scanf("%I64d%I64d",&x,&k); lll val = (x + d[u]*k)%Mod; modify(0,L[u],val); modify(0,R[u]+1,-val); modify(1,L[u],-k); modify(1,R[u]+1,k); } else { lll ans = ((getsum(0,L[u])+getsum(1,L[u])*d[u])%Mod+Mod)%Mod; printf("%I64d\n",ans); } } } return 0; }