181. [USACO Jan07] 最高的牛
★★ 输入文件：tallest.in 输出文件：tallest.out 简单对比
时间限制：1 s 内存限制：32 MB
FJ‘s N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.
FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form "cow 17 sees cow 34". This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.
For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.
为了方便我们把FJ养的N头(1 ≤ N ≤ 10,000)奶牛从1号开始编号，一直编到N号，并且让他们站成一列。每头牛都有一个正整数用来描述她的身高(有些牛不愿意说，她们的身高保密)。你只知道这些牛中最高的那一头的身高是H，她的编号是第I号。
FJ写了一张单子，上面有R个陈述句，大概是”第17号牛可以看见第34号牛”这种。这意味着第34号牛最起码和第17号牛一样高。而且17号牛和34号牛之间的牛得身高都严格的小于min(H[17],H[34])。
从第1号牛到第N号牛，求她们可能拥有的最大身高，所给出的所有信息都是正确的。这可以保证满足所有条件。
Input
- Line 1: Four space-separated integers: N, I, H and R
- Lines 2..R + 1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.
输入：
第1行：四个由空格分开的整数，N，I，H和R。
第2..R+1行：两个由空格分开的整数A和B(1 ≤ A, B ≤ N),表示第A号牛可以看到第B号牛。
Output
- Lines 1..N: Line i contains the maximum possible height of cow i.
输出：
第1..N行:第i行包括第i号牛可能拥有的最大身高。
Sample Input
9 3 5 5 1 3 5 3 4 3 3 7 9 8
Sample Output
5 4 5 3 4 4 5 5 5
Input Details
There are 9 cows, and the 3rd is the tallest with height 5.
输入细节：
由9头奶牛，第三头奶牛最高，身高是5.
- 译byKZFFFFFFFF
思路：线段树。
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 10000
using namespace std;
map<int,int>ma[MAXN];
struct nond{
int l,r,flag;
}tree[MAXN*4];
long long n,I,H,R;
void build(int now,int l,int r){
tree[now].l=l;tree[now].flag=0;
tree[now].r=r;
if(tree[now].l==tree[now].r)
return ;
long long mid=(tree[now].l+tree[now].r)/2;
build(now*2,l,mid);
build(now*2+1,mid+1,r);
}
void down(int now){
tree[now*2].flag+=tree[now].flag;
tree[now*2+1].flag+=tree[now].flag;
tree[now].flag=0;
}
void change(int now,int l,int r){
if(tree[now].l==l&&tree[now].r==r){
tree[now].flag+=1;
return ;
}
if(tree[now].flag) down(now);
int mid=(tree[now].l+tree[now].r)/2;
if(r<=mid) change(now*2,l,r);
else if(l>mid) change(now*2+1,l,r);
else{
change(now*2,l,mid);
change(now*2+1,mid+1,r);
}
}
int query(int now,int x){
if(tree[now].l==tree[now].r)
return tree[now].flag;
if(tree[now].flag) down(now);
int mid=(tree[now].l+tree[now].r)/2;
if(x<=mid) return query(now*2,x);
else if(x>mid) return query(now*2+1,x);
}
int main(){
//freopen("tallest.in","r",stdin);
//freopen("tallest.out","w",stdout);
while(scanf("%lld%lld%lld%lld",&n,&I,&H,&R)!=EOF){
build(1,1,n);
for(int i=1;i<=R;i++){
long long a,b;
scanf("%lld%lld",&a,&b);
if(a>b) swap(a,b);
if(ma[a][b]) continue;
ma[a][b]=1;
if(a+1==b) continue;
change(1,a+1,b-1);
}
for(long long i=1;i<=n;i++)
cout<<H-query(1,i)<<endl;
}
}
一直调不对的20分代码
AC代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=10000+100;
int LL[maxn],RR[maxn];
long long N,H,R,I;
long long add[maxn<<2],ans[maxn];
struct node{
int l,r;
int sum;
}t[maxn<<2];
void pushdown(int rs)//向下更新lazy积累的值
{
if(add[rs])
{
add[rs<<1]+=add[rs];
add[rs<<1|1]+=add[rs];
add[rs]=0;
}
}
void build(int rs,int l,int r)
{
t[rs].l=l;
t[rs].r=r;
add[rs]=0;
if(l==r)
{
t[rs].sum=H;
return ;
}
int mid=(l+r)>>1;
build(rs<<1,l,mid);
build(rs<<1|1,mid+1,r);
}
void update(int l,int r,int rs)//更新操作
{
if(t[rs].l>=l&&t[rs].r<=r)
{
add[rs]+=1;
return ;
}
pushdown(rs);
int mid=(t[rs].l+t[rs].r)>>1;
if(l<=mid) update(l,r,rs<<1);
if(r>mid) update(l,r,rs<<1|1);
}
int query(int k,int rs)
{
if(t[rs].l==t[rs].r)
{
return t[rs].sum-=add[rs];
}
pushdown(rs);
int mid=(t[rs].l+t[rs].r)>>1;
if(k<=mid) return query(k,rs<<1);
else return query(k,rs<<1|1);
}
int main()
{
freopen("tallest.in","r",stdin);
freopen("tallest.out","w",stdout);
int u,v;
scanf("%I64d%I64d%I64d%I64d",&N,&I,&H,&R);
build(1,1,N);
int k=0;
while(R--)
{
scanf("%d%d",&u,&v);
if(u>v) swap(u,v);
int flag=1;
for(int i=0;i<k;i++)//判断重边，在poj做题一定要防这东西
{
if(u==LL[i]&&v==RR[i])
{
flag=0;
break;
}
}
if(!flag)
continue;
LL[k]=u;RR[k]=v;
k++;
if(v-u==1)//相邻的话就不用更新了
continue;
update(u+1,v-1,1);//注意是开区间的值更新
}
for(int i=1;i<=N;i++)
ans[i]=query(i,1);
for(int i=1;i<=N;i++)
printf("%I64d\n",ans[i]);
return 0;
}
原文地址：https://www.cnblogs.com/cangT-Tlan/p/8151173.html