参考答案
感觉这两题目都挺好的~~
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;
}