Codeforces Round #303 (Div. 2)——A.B.C.D.E

http://codeforces.com/contest/545

好水的一场div2.
A. Toy Cars
a[i][j] = 1，表示第i辆车坏掉了，=2 表示第j辆车坏掉了，=3表示i和j都坏掉了。求有几辆车是好的

#include <bits/stdc++.h>#define foreach(v,i) for(__typeof((v).begin()) i=(v).begin();i!=(v).end();++i)const int MAXN = 110;using namespace std;vector<int> vi;bool vis[MAXN];int n;int main(){#ifndef ONLINE_JUDGEfreopen("in.cpp","r",stdin);freopen("out.cpp","w",stdout);#endif // ONLINE_JUDGE    int x;    scanf("%d",&n);    for(int i=1;i<=n;++i){        for(int j=1;j<=n;++j){            scanf("%d",&x);            if(x==1) vis[i]=true;            if(x==2) vis[j]=true;            if(x==3){                vis[i]=vis[j]=true;            }        }    }    int res=0;    for(int i=1;i<=n;++i){        res+=vis[i];        if(!vis[i]) vi.push_back(i);    }    cout<<n-res<<endl;    foreach(vi,i){        cout<<*i<<" ";    }    return 0;}

B. Equidistant String
给两个由0,1构成的a,b串，求c串使得c到a的距离=c到b的距离。
距离为每个位置上数字差的绝对值的和

#include<bits/stdc++.h>const int MAXN = 100010;using namespace std;char a[MAXN],b[MAXN];int main(){#ifndef ONLINE_JUDGEfreopen("in.cpp","r",stdin);freopen("out.cpp","w",stdout);#endif // ONLINE_JUDGE    scanf("%s%s",a,b);    int n=strlen(a);    int cnt=0;    for(int i=0;i<n;++i){        if(a[i]!=b[i]) cnt++;    }    if(cnt&1){        puts("impossible");        return 0;    }    string s;    cnt=0;    for(int i=0;i<n;++i){        if(a[i]!=b[i]){            if(cnt&1) s+=a[i];            else s+=b[i];            cnt++;        }        else s+=a[i];    }    cout<<s<<endl;    return 0;}

C. Woodcutters
n棵树并排在同一排，每棵树有个横坐标和高度，伐木工人砍树可以让树向左倒或者向右倒（必须有足够的区间长度),求最多可以砍倒多少棵树

最左边的树一定要向左倒，结果不会更差，然后对于每一棵树，能砍就砍

#include <bits/stdc++.h>const int MAXN = 100010;const int INF = 0x3f3f3f3f;using namespace std;int x[MAXN],h[MAXN];int len[MAXN];int n;int main(){#ifndef ONLINE_JUDGEfreopen("in.cpp","r",stdin);freopen("out.cpp","w",stdout);#endif // ONLINE_JUDGE    scanf("%d",&n);    x[0]=0;    for(int i=1;i<=n;++i){        scanf("%d%d",&x[i],&h[i]);        len[i]=x[i]-x[i-1];    }    int cnt=n>=2 ? 2:1;    for(int i=2;i<=n-1;++i){        if(len[i]>h[i]) cnt++;        else if(len[i+1]>h[i]){            len[i+1]-=h[i];            cnt++;        }    }    cout<<cnt<<endl;    return 0;}

D. Queue
n个人在超市排队，每个人都有一个服务时间，当一个人的等待时间大于服务时间，这个人就不买东西了。然后可以随意交换每个人的位置，求最多可以让多少个人买东西

排个序，枚举就行了

#include <bits/stdc++.h>typedef long long LL;const int MAXN = 100010;using namespace std;int n;LL t[MAXN];int main(){#ifndef ONLINE_JUDGE    freopen("in.cpp","r",stdin);    freopen("out.cpp","w",stdout);#endif // ONLINE_JUDGE    while(scanf("%d",&n)==1) {        for(int i=1; i<=n; ++i) {            scanf("%I64d",&t[i]);        }        sort(t+1,t+n+1);        LL sum=0;        int cnt=0;        for(int i=1; i<=n; ++i) {            if(sum>t[i]){                cnt++;            }else sum+=t[i];        }        cout<<n-cnt<<endl;    }    return 0;}

E. Paths and Trees
一个带权无向图，求一个新图G’=(V,E’)，使得源点s到新图各个点的最短距离等于在原图中的最短距离，输出边权值最小的新图

dijkstra，求一次单源最短路，在距离相同的情况下，维护边权值小的

#include <bits/stdc++.h>#define LL long long#define pii pair<int,int>#define mk make_pair#define clr(a,b) memset(a,b,sizeof(a))#define foreach(v,i) for(__typeof((v).begin()) i=(v).begin();i!=(v).end();++i)const int MAXN = 300010;const LL INF = ~0uLL>>1;using namespace std;set<LL> si;LL d[MAXN];int p[MAXN];bool vis[MAXN];int n,m;struct HeapNode{    LL d;    int u;    bool operator<(const HeapNode& rhs)const{        return d>rhs.d;    }};struct Edge{    int to,next;    LL w;    int id;}edge[MAXN<<1];int head[MAXN],tot;void init(){    tot=0;    clr(head,0xff);}void addedge(int u,int v,LL w,int id){    edge[tot].to=v;    edge[tot].w=w;    edge[tot].id=id;    edge[tot].next=head[u];    head[u]=tot++;}void dijkstra(int s){    priority_queue<HeapNode> q;    for(int i=1;i<=n;++i) d[i]=INF;    d[s]=0;    clr(vis,false);    clr(p,0xff);    q.push((HeapNode){0,s});    while(!q.empty()){        HeapNode x=q.top();q.pop();        int u=x.u;        if(vis[u]) continue;        vis[u]=1;        for(int i=head[u];i!=-1;i=edge[i].next){            int v=edge[i].to;            LL w=edge[i].w;            if(d[v]>d[u]+w){                d[v]=d[u]+w;                p[v]=i;                q.push((HeapNode){d[v],v});            }            else if(d[v]==d[u]+w&&edge[p[v]].w>w){                p[v]=i;            }        }    }}void print(int s){    LL ans=0;    for(int i=1;i<=n;++i){        if(p[i]>=0) ans+=edge[p[i]].w;    }    cout<<ans<<endl;    for(int i=1;i<=n;++i){        if(p[i]>=0) si.insert(edge[p[i]].id);    }    foreach(si,i){        printf("%d ",*i);    }    puts("");}int main(){#ifndef ONLINE_JUDGEfreopen("in.cpp","r",stdin);freopen("out.cpp","w",stdout);#endif // ONLINE_JUDGE    int u,v,s;    LL w;    scanf("%d%d",&n,&m);    init();    for(int i=1;i<=m;++i){        scanf("%d%d%I64d",&u,&v,&w);        addedge(u,v,w,i);        addedge(v,u,w,i);    }    scanf("%d",&s);    dijkstra(s);    print(s);    return 0;}