CodeForces701E DFS

其他 2018-08-19 09:47:47 阅读次数: 0

http://codeforces.com/problemset/problem/701/E

一个显而易见的方法是考虑点的贡献，一次dfs记录到所有根节点不考虑匹配的答案，再一次dfs反向推出答案 

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1;
inline bool isdigit(const char& n) {return (n>='0'&&n<='9');}
inline void rnext(){if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);}
template <class T> inline bool in(T &ans){
#ifdef VSCode
ans=0;T f=1;register char c;
do{c=getchar();if ('-'==c)f=-1;}while(!isdigit(c)&&c!=EOF);
if(c==EOF)return false;do{ans=(ans<<1)+(ans<<3)+c-48;
c=getchar();}while(isdigit(c)&&c!=EOF);ans*=f;return true;
#endif
#ifndef VSCode 
ans =0;T f=1;if(ps==pe)return false;do{rnext();if('-'==*ps)f=-1;} 
while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(ans<<1)+(ans<<3)+*ps-48;
rnext();}while(isdigit(*ps)&&ps!=pe);ans*=f;return true;
#endif
}const int MAXOUT=10000;
char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
inline void write(){fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;}
inline void out_char(char c){*(pout++)=c;if(pout==pend)write();}
inline void out_str(char *s){while(*s){*(pout++)=*(s++);if(pout==pend)write();}}
template <class T>inline void out_int(T x) {if(!x){out_char('0');return;}
if(x<0)x=-x,out_char('-');int len=0;while(x){outtmp[len++]=x%10+48;x/=10;}outtmp[len]=0;
for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp);}
template<typename T, typename... T2>
inline int in(T& value, T2&... value2) { in(value); return in(value2...); }
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
#define Vec Point
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,K,temp,cnt;
bool sp[maxn];
LL head[maxn],MAX[maxn],id[maxn],sum[maxn],ans[maxn];
struct Edge{
    int to,next;
}edge[maxn * 2];
void init(){Mem(head,0); cnt = 0;}
void add(int u,int v){edge[++cnt].to = v; edge[cnt].next = head[u];head[u] = cnt;}
LL dfs(int x,int last){
    sum[x] = 0;
    for(int i = head[x] ; i; i = edge[i].next){
        int v = edge[i].to;
        if(v == last) continue;
        int t = dfs(v,x);
        if(t > MAX[x]){
            MAX[x] = t;
            id[x] = v;
        }
        sum[x] += t;
        ans[x] += ans[v];
    }
    ans[x] += sum[x];
    sum[x] += sp[x];
    return sum[x];
}
LL getans(int x,int c){
    sum[x] -= c;
    MAX[x] -= c;
    if(MAX[x] * 2 <= sum[x]){
        return ans[x];
    }else{
        int v = id[x];
        return ans[x] - ans[v] - sum[v] + getans(v,sum[x] - MAX[x] + c) + sum[x] - MAX[x] + c;
    }
}
LL solve(){
    int root = 1;
    dfs(root,-1);
    // For(i,1,N){
    //     cout << i << ":" << sum[i] << " " << MAX[i] << endl;
    // }
    return getans(root,0);
}
int main()
{
    in(N,K);
    For(i,1,2 * K){
        in(temp);
        sp[temp] = 1;
    }
    For(i,1,N - 1){
        int u,v; in(u,v);
        add(u,v); add(v,u);
    }
    Prl(solve());
    #ifdef VSCode
    write();
    system("pause");
    #endif
    return 0;
}
View Code

这并不是很好写，另一个不显而易见但是很好写很好对的方法是考虑边的贡献，一条边两边的点分别是x和2K - x,由于路线尽可能的长，较小的那一边一定要加上较大的那一边，所以每条边经过一次，同时更新ans即可

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
const int MAXBUF=10000;char buf[MAXBUF],*ps=buf,*pe=buf+1;
inline bool isdigit(const char& n) {return (n>='0'&&n<='9');}
inline void rnext(){if(++ps==pe)pe=(ps=buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);}
template <class T> inline bool in(T &ans){
#ifdef VSCode
ans=0;T f=1;register char c;
do{c=getchar();if ('-'==c)f=-1;}while(!isdigit(c)&&c!=EOF);
if(c==EOF)return false;do{ans=(ans<<1)+(ans<<3)+c-48;
c=getchar();}while(isdigit(c)&&c!=EOF);ans*=f;return true;
#endif
#ifndef VSCode 
ans =0;T f=1;if(ps==pe)return false;do{rnext();if('-'==*ps)f=-1;} 
while(!isdigit(*ps)&&ps!=pe);if(ps==pe)return false;do{ans=(ans<<1)+(ans<<3)+*ps-48;
rnext();}while(isdigit(*ps)&&ps!=pe);ans*=f;return true;
#endif
}const int MAXOUT=10000;
char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
inline void write(){fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;}
inline void out_char(char c){*(pout++)=c;if(pout==pend)write();}
inline void out_str(char *s){while(*s){*(pout++)=*(s++);if(pout==pend)write();}}
template <class T>inline void out_int(T x) {if(!x){out_char('0');return;}
if(x<0)x=-x,out_char('-');int len=0;while(x){outtmp[len++]=x%10+48;x/=10;}outtmp[len]=0;
for(int i=0,j=len-1;i<j;i++,j--) swap(outtmp[i],outtmp[j]);out_str(outtmp);}
template<typename T, typename... T2>
inline int in(T& value, T2&... value2) { in(value); return in(value2...); }
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
#define Vec Point
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,K,temp,cnt;
bool sp[maxn];
LL head[maxn],MAX[maxn],id[maxn],sum[maxn],ans[maxn],Ans;
struct Edge{
    int to,next;
}edge[maxn * 2];
void init(){Mem(head,0); cnt = 0;}
void add(int u,int v){edge[++cnt].to = v; edge[cnt].next = head[u];head[u] = cnt;}
LL dfs(int x,int last){
    sum[x] = sp[x];
    for(int i = head[x] ; i; i = edge[i].next){
        int v = edge[i].to;
        if(v == last) continue;
        int t = dfs(v,x);
        Ans += min(t,2 * K - t);
        sum[x] += t;
    }
    return sum[x];
}
LL solve(){
    int root = 1;
    dfs(root,-1);
    return Ans;
}
int main()
{
    in(N,K);
    For(i,1,2 * K){
        in(temp);
        sp[temp] = 1;
    }
    For(i,1,N - 1){
        int u,v; in(u,v);
        add(u,v); add(v,u);
    }
    Prl(solve());
    #ifdef VSCode
    write();
    system("pause");
    #endif
    return 0;
}

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转载自www.cnblogs.com/Hugh-Locke/p/9499721.html
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