kuangbin专题七 ZOJ1610 Count the Colors （灵活线段树）

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

Input

The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

 

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

<b< dd="">

Output

Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.

 

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

<b< dd="">

Sample Input

5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1

 

<b< dd="">

Sample Output

1 1
2 1
3 1

 

1 1

0 2
1 1

 

因为是线段树专题，就思考线段树，由于刚用线段树，这里的建树感觉没用，但是不确定，不过Pushup果断弃用。还有就是Query的时候是需要跑所有叶节点的，所以不需要那些判断， 大体思路正确，就是没想到用last记录。

 

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define FO freopen("in.txt","r",stdin);
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define debug(x) cout << "&&" << x << "&&" << endl;
#define lowbit(x) (x&-x)
#define mem(a,b) memset(a, b, sizeof(a));
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
const int inf = 0x3f3f3f3f;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//head

const int maxn=8010;                //last 上一段的颜色
int lazy[maxn<<2],n,ans[maxn],last;//lazy就是树，不需要处理什么信息 

void Pushdown(int rt) {
    if(lazy[rt]!=-1) {
        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
        lazy[rt]=-1;
    }
}

void Updata(int rt,int L,int R,int l,int r,int val) {
    if(L>=l&&R<=r) {
        lazy[rt]=val;
        return;
    }
    Pushdown(rt);
    int mid=(L+R)>>1;
    if(l<=mid) Updata(rt<<1,L,mid,l,r,val);
    if(r>mid) Updata(rt<<1|1,mid+1,R,l,r,val);
}

void Query(int rt,int L,int R,int l,int r) {//不要太死板，学会变通
    if(L==R) {
        if(lazy[rt]!=-1&&lazy[rt]!=last) {//巧妙
            ans[lazy[rt]]++;
        }
        last=lazy[rt];
        return;
    }
    Pushdown(rt);
    int mid=(L+R)>>1;//需要遍历所有叶节点
    Query(rt<<1,L,mid,l,r);
    Query(rt<<1|1,mid+1,R,l,r);
}

int main() {
    while(~scanf("%d",&n)) {
        mem(ans,0);
        mem(lazy,-1);
        int x,y,val;
        while(n--) {
            scanf("%d%d%d",&x,&y,&val);
            Updata(1,1,maxn,x+1,y,val);//把0 变为 1
        }
        last=-1;
        Query(1,1,maxn,1,maxn);
        rep(i,0,maxn) {
            if(ans[i]) printf("%d %d\n",i,ans[i]);
        }
        printf("\n");
    }
}