codeforces contest 357

http://codeforces.com/contest/357

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C 356A Knight
Tournament

题意：n个人，m场比赛。下面m行每行三个数l,r,x。表示x打败了l到r

题解：

解法一：丢进set，二分查找l，然后删除l到r。复杂度O(mlogn)

解法二：线段树倒着做。区间修改单点查询。复杂度O(mlogn)

解法三：并查集。（刷完并查集专题再来做）

#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define de(x) cout << #x << "=" << x << endl
const int N=300005;
set<int> s;
int ans
,tmp
;
int main() {
int n,m;
while(~scanf("%d%d",&n,&m)) {
s.clear();
for(int i=1;i<=n;++i) s.insert(i);
memset(ans,0,sizeof(ans));
int l,r,x;
for(int i=1;i<=m;++i) {
scanf("%d%d%d",&l,&r,&x);
int cnt=0;
for(set<int>::iterator it=s.lower_bound(l);it!=s.end();++it) {
if(*it>r) break;
if(*it==x) continue;
tmp[++cnt]=*it;
ans[*it]=x;
}
for(int j=1;j<=cnt;++j) s.erase(tmp[j]);
}
for(int i=1;i<=n;++i) printf("%d ",ans[i]);puts("");
}
return 0;
}

#include<cstdio>
#include<cstring>
const int N=300005;
struct Node {
int l,r,x;
}a
;
int tree[N<<2],lazy[N<<2];
void init() {
memset(tree,0,sizeof(tree));
memset(lazy,-1,sizeof(lazy));
}
void pushdown(int now) {
if(lazy[now]==-1) return ;
tree[now<<1]=tree[now<<1|1]=lazy[now<<1]=lazy[now<<1|1]=lazy[now];
lazy[now]=-1;
}
void update(int L,int R,int val,int l,int r,int now) {//把[L,R]的值变成val
if(L<=l&&r<=R) {
tree[now]=lazy[now]=val;
return ;
}
pushdown(now);
int mid=l+r>>1;
if(mid>=L) update(L,R,val,l,mid,now<<1);
if(mid+1<=R) update(L,R,val,mid+1,r,now<<1|1);
}
int query(int p,int l,int r,int now) {
if(l==r) return tree[now];
pushdown(now);
int mid=l+r>>1;
if(p<=mid) return query(p,l,mid,now<<1);
else return query(p,mid+1,r,now<<1|1);
}
int main() {
int n,m;
while(~scanf("%d%d",&n,&m)) {
for(int i=1;i<=m;++i) scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].x);
init();
for(int i=m;i>=1;--i) {
if(a[i].l<=a[i].x-1) update(a[i].l,a[i].x-1,a[i].x,1,n,1);
if(a[i].x+1<=a[i].r) update(a[i].x+1,a[i].r,a[i].x,1,n,1);
}
for(int i=1;i<=n;++i) printf("%d ",query(i,1,n,1));
puts("");
}
return 0;
}

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D 356B Xenia
and Hamming
题意：s1重复出现n次得到str1，s2重复出现m次得到str2，len==len(str1)==len(str2)。求有多少个i(1~len)，使得str1[i]!=str2[i]
题解：处理出LCM再去做肯定会超时。具体见代码。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define de(x) cout << #x << "=" << x << endl
const int N=1000005;
char s1
,s2
;
ll gcd(ll a,ll b) {
if(b==0) return a;
return gcd(b,a%b);
}
int cnt
[30];
int main() {
ll n,m;
while(~scanf("%I64d%I64d%s%s",&n,&m,s1,s2)) {
int len1=strlen(s1);
int len2=strlen(s2);
int d=gcd(len1,len2);
memset(cnt,0,sizeof(cnt));
for(int i=0;i<len1;++i) ++cnt[i%d][s1[i]-'a'];
ll ans=0;
for(int i=0;i<len2;++i) ans+=cnt[i%d][s2[i]-'a'];
ll len=1ll*len1*len2/d;
ans=len-ans;
ans=ans*(n*len1/len);
//ans=n*len1-ans*n*d/len2;
printf("%I64d\n",ans);
}
return 0;
}

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E 356C Compartments
题意：n个车厢每个车厢a[i]个学生（最多四个）。让最少的学生和别人交换座位，使得每个车厢学生数为0或3或4。

题解：贪心。可以考虑各种方案中满足x个车厢的花费。具体见代码。

#include<cstring>
#include<cstdio>
int a[5];
int main() {
int n,x;
while(~scanf("%d",&n)) {
memset(a,0,sizeof(a));
for(int i=1;i<=n;++i) {
scanf("%d",&x);
++a[x];
}
int ans=0;
if(a[1]<=a[2]) {
//1 2-->3
a[2]-=a[1];
a[3]+=a[1];
ans+=a[1];
//2 2 2-->3 3
int t=a[2]/3;
a[2]%=3;
a[3]+=t*2;
ans+=t*2;
if(a[2]==1) {
//2 4-->3 3
if(a[4]) {
++ans;
//2 3 3-->4 4
} else if(a[3]>=2) {
ans+=2;
} else {
ans=-1;
}
//2 2-->4
} else if(a[2]==2) {
ans+=2;
}
} else {
//1 2-->3
a[1]-=a[2];
a[3]+=a[2];
ans+=a[2];
//1 1 1-->3
int t=a[1]/3;
a[1]%=3;
a[3]+=t;
ans+=t*2;
if(a[1]==1) {
//1 3-->4
if(a[3]) {
++ans;
//1 4 4-->3 3 3
} else if(a[4]>=2) {
ans+=2;
} else {
ans=-1;
}
} else if(a[1]==2) {
//1 1 4-->3 3
if(a[4]) {
ans+=2;
//1 1 3 3-->4 4
} else if(a[3]>=2) {
ans+=2;
} else {
ans=-1;
}
}
}
printf("%d\n",ans);
}
return 0;
}