Ural 1019 Line Painting

点击打开链接ural 1019

思路：离散化

分析：

1 这一题的区间的最大值为10^9而n最大为5000，很明显就是利用离散化

2 题目中说了区间[0,10^9]刚开始为白色，而给定重刷的区间的值是大于0小于10^9的，所以我们应该开始就要考虑到0和10^9.

3 然后我们应该注意这题是对线段染色，所以我们应该要注意在处理的时候应该要把这些区间看成点来处理，然后会有两种方法分别是暴力和线段树

代码

线段树

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 10010;
struct Node{
int left;
int right;
int mark;
char color;
};
Node node[4*MAXN] , tmp[MAXN];
int n , pos;
int num[MAXN];

//二分查找
int search(int x){
int left , right;
left = 1 , right = pos-1;
while(left <= right){
int mid = (left + right)>>1;
if(num[mid] == x)
return mid;
else if(num[mid] > x)
right = mid-1;
else
left = mid+1;
}
}

//建立线段树
void buildTree(int left , int right , int pos){
node[pos].left = left;
node[pos].right = right;
node[pos].mark = 0;
if(left == right)
return;
int mid = (left+right)>>1;
buildTree(left , mid , pos<<1);
buildTree(mid+1 , right , (pos<<1)+1);
}

//向下更新
void push_down(int pos){
if(node[pos].mark != -1){
node[pos<<1].mark = node[(pos<<1)+1].mark = node[pos].mark;
node[pos].mark = -1;
}
}

//更新
void update(int left , int right , int value , int pos){
if(left <= node[pos].left && right >= node[pos].right){
node[pos].mark = value;
return;
}
push_down(pos);
int mid = (node[pos].left + node[pos].right)>>1;
if(right <= mid)
update(left , right , value , pos<<1);
else if(left > mid)
update(left , right , value , (pos<<1)+1);
else{
update(left , mid , value , pos<<1);
update(mid+1 , right , value , (pos<<1)+1);
}
}

//询问
int query(int index , int pos){
if(node[pos].left == node[pos].right)
return node[pos].mark;
push_down(pos);
int mid = (node[pos].left + node[pos].right)>>1;
if(index <= mid)
return query(index , pos<<1);
else
return query(index , (pos<<1)+1);
}

int main(){
while(scanf("%d" , &n) != EOF){
pos = 1;
num[pos++] = 0;
num[pos++] = 1e9;
for(int i = 0 ; i < n ; i++){
scanf("%d %d %c" , &tmp[i].left , &tmp[i].right , &tmp[i].color);
num[pos++] = tmp[i].left;
num[pos++] = tmp[i].right;
}
sort(num+1 , num+pos);
pos = unique(num+1 , num+pos)-num;
//离散化
for(int i = 0 ; i < n ; i++){
tmp[i].left = search(tmp[i].left);
tmp[i].right = search(tmp[i].right);
}
buildTree(1 , pos-1 , 1);
//更新
for(int i = 0 ; i < n ; i++){
int value = 0;
if(tmp[i].color == 'b')
value = 1;
update(tmp[i].left , tmp[i].right-1 , value , 1);
}
//求解最长的白色区间
int left , right;
left = right = 0;
for(int i = 1 ; i < pos-1 ; i++){
if(!query(i , 1)){
int j = i+1;
while(j < pos-1 && !query(j , 1))
j++;
int dis = num[j]-num[i];
if(dis > right-left){
left = num[i];
right = num[j];
}
i = j;
}
}
printf("%d %d\n" , left , right);
}
return 0;
}


暴力

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 100010;
struct Node{
int left;
int right;
int mark;
char color;
};
Node node[MAXN];
int vis[MAXN];
int num[MAXN];
int pos;

//二分查找
int search(int x){
int left = 1 , right = pos-1;
while(left <= right){
int mid = (left+right)>>1;
if(num[mid] == x)
return mid;
else if(num[mid] > x)
right = mid-1;
else
left = mid+1;
}
}

int main(){
int n;
while(scanf("%d" , &n) != EOF){
pos = 1;
num[pos++] = 0;
num[pos++] = 1e9;
for(int i = 0 ; i < n ; i++){
scanf("%d %d %c" , &node[i].left , &node[i].right , &node[i].color);
num[pos++] = node[i].left;
num[pos++] = node[i].right;
}
sort(num+1 , num+pos);
pos = unique(num+1 , num+pos)-num;
for(int i = 0 ; i < n ; i++){
node[i].left = search(node[i].left);
node[i].right = search(node[i].right);
}
//染色
memset(vis , 0 , sizeof(vis));
for(int i = 0 ; i < n ; i++){
int value = 0;
if(node[i].color == 'b')
value = 1;
for(int j = node[i].left ; j < node[i].right ; j++)
vis[j] = value;
}
int left , right;
left = right = 0;
//找最大的连续的白色区间
for(int i = 1 ; i < pos-1 ; i++){
if(!vis[i]){
int j = i+1;
while(j < pos-1 && !vis[j])
j++;
int dis = num[j]-num[i];
if(dis > right-left){
right = num[j];
left = num[i];
}
i = j;
}
}
printf("%d %d\n" , left , right);
}
return 0;
}