【NOIP 2016】初赛-完善程序 & 参考答案

参考答案

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 感觉这两题目都挺好的~~

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T1 交朋友

简单描述:有n个人依次进入教室,每个人进入会找一个身高绝对值相差最小的人交朋友(相同时更想和高的交朋友),求每个人交的朋友.

Solution:

Sort,求出每个人的排名

逆向思维,从后往前(每次删除,然后剩余的都是可以选的)

链表存储前一个和后一个,每次删除发生改变

发组福利数据

in:

6
4 6 5 3 1 7

out:

2:1
3:2
4:1
5:4
6:2

// <T1.cpp> - Sun Oct 23 22:05:36 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define INF 0x7fffffff
using namespace std;
const int MAXN=100010;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
int as[MAXN],rank[MAXN],h[MAXN],pre[MAXN],nex[MAXN];
bool cmp(int a,int b){return h[a]<h[b];}
int main()
{
	freopen("T1.in","r",stdin);
	freopen("T1.out","w",stdout);
	int n=gi(),s;
    for(int i=1;i<=n;i++)h[i]=gi(),rank[i]=i;
    sort(rank+1,rank+1+n,cmp);
    for(int i=1;i<=n;i++)
        pre[rank[i]]=rank[i-1],nex[rank[i]]=rank[i+1];
    for(int i=n;i>=2;i--){
        s=INF;
        if(!pre[i])s=h[i]-h[pre[i]];
        if(!nex[i])s+=h[i]-h[nex[i]];
        if(s<0)as[i]=pre[i];
        else as[i]=nex[i];
        nex[pre[i]]=nex[i];
        pre[nex[i]]=pre[i];
    }
    for(int i=2;i<=n;i++)printf("%d:%d
",i,as[i]);
	return 0;
}

T2 交通中断

大意:给一个无向图,问第x个点中断时(与其他点连边全部删掉),从1号点到多少个点的最短路改变(包括不能到达，不包括x)

Solution:

先求出1到每个点的最短路,然后举删掉了哪个点,从1开始bfs(到达的是最短路,加入队列),这样贪心保证正确性,因为一个点可能可以从多个点跑相同长度的最短路到达.

发组福利数据

in:

5 7
1 2 1
1 2 0
1 3 4
2 3 3
4 5 1
3 4 2
1 4 3

out:

2:1
3:0
4:1
5:0

// <T2.cpp> - Sun Oct 23 22:05:36 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#define IN inline
#define RG register
#define INF 0x7fffffff
using namespace std;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
struct SPFA{
    static const int N=6010,M=100010;
    queue<int>q;bool u[N];
    int n,m,t;int d[N],fr[N];int to[M],ne[M],W[M];
    IN void link(RG int u,RG int v,RG int w){
        to[++t]=v;ne[t]=fr[u];fr[u]=t;W[t]=w;
    }
    void read(){
        n=gi(),m=gi();
        while(m--){
            int u=gi(),v=gi(),w=gi();
            link(u,v,w);link(v,u,w);
        }
    }
    void Spfa(int begin){
        for(int i=1;i<=n;i++)d[i]=INF;
        q.push(begin);d[begin]=0;memset(u,0,sizeof(u));
        while(!q.empty()){
            int x=q.front();q.pop();u[x]=0;
            for(int o=fr[x],y;y=to[o],o;o=ne[o])
                if(d[x]+W[o]<d[y]){
                    d[y]=d[x]+W[o];
                    if(!u[y])u[y]=1,q.push(y);
                }
        }
    }
    void Work(){
        read();Spfa(1);
        while(!q.empty())q.pop();
        for(int i=2;i<=n;i++){
            memset(u,0,sizeof(u));
            q.push(1);u[1]=1;
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int o=fr[x],y;y=to[o],o;o=ne[o])
                    if(y!=i&&d[x]+W[o]==d[y]&&!u[y])
                        u[y]=1,q.push(y);
            }int ans=0;
            for(int j=1;j<=n;j++)ans+=1-u[j];
            printf("%d:%d
",i,ans-1);
        }
    }
}e;
int main()
{
	freopen("T2.in","r",stdin);
	freopen("T2.out","w",stdout);
	e.Work();
	return 0;
}