Count the Colors

B - Count the Colors

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

Description

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.

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.

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

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

/*Author: 2486Memory: 232 KBTime: 110 MSLanguage: C++ (g++ 4.7.2)Result: Accepted*///暴力可过，或者是线段树#include <cstdio>#include <cstring>#include <algorithm>typedef long long ll;using namespace std;const int maxn=8000+5;int n,a,b,c,ans[maxn],col[maxn];int main() {    while(~scanf("%d",&n)) {        memset(ans,0,sizeof(ans));        memset(col,0,sizeof(col));        int Max=0;        for(int i=0; i<n; i++) {            scanf("%d%d%d",&a,&b,&c);            for(int j=a; j<b; j++) {                col[j]=c+1;            }            Max=max(Max,b);        }        for(int i=0; i<=Max; i++) {            while(i!=0&&col[i]&&col[i]==col[i-1]) {                i++;            }            if(col[i]) {                ans[col[i]-1]++;            }        }        for(int i=0; i<=8001; i++) {            if(ans[i]) {                printf("%d %d\n",i,ans[i]);            }        }        printf("\n");    }}

/*Author: 2486Memory: 356 KBTime: 20 MSLanguage: C++ (g++ 4.7.2)Result: Accepted*/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=8000+5;int sum[maxn],col[maxn<<2],res[maxn];int n,x1,x2,c;void pushup(int rt) {}void pushdown(int rt,int l,int r) {    if(col[rt]!=-1) {        col[rt<<1]=col[rt<<1|1]=col[rt];//直接往下递归就可以了，不用管其他        col[rt]=-1;    }}void build(int rt,int l,int r) {    col[rt]=-1;    if(l==r)return;    int mid=(l+r)>>1;    build(rt<<1,l,mid);    build(rt<<1|1,mid+1,r);}void update(int L,int R,int c,int rt,int l,int r) {    if(L<=l&&r<=R) {        col[rt]=c;        return;    }    pushdown(rt,l,r);    int mid=(l+r)>>1;    if(L<=mid)update(L,R,c,rt<<1,l,mid);    if(mid<R)update(L,R,c,rt<<1|1,mid+1,r);    pushup(rt);}void query(int L,int R,int rt,int l,int  r) {    if(l==r) {        sum[l]=col[rt];//赋值，放回区间的位子颜色        return;    }    pushdown(rt,l,r);    int mid=(l+r)>>1;    if(L<=mid)query(L,R,rt<<1,l,mid);    if(mid<R)query(L,R,rt<<1|1,mid+1,r);    pushup(rt);}int main() {    while(~scanf("%d",&n)) {        memset(sum,-1,sizeof(sum));        memset(res,0,sizeof(res));        build(1,0,maxn-1);        for(int i=0; i<n; i++) {            scanf("%d%d%d",&x1,&x2,&c);            if(x1==x2)continue;            update(x1,x2-1,c,1,0,maxn-1);        }        query(0,maxn-1,1,0,maxn-1);//将所有数据都更新到底部        for(int i=0; i<maxn; i++) {            while(i!=0&&sum[i]!=-1&&sum[i]==sum[i-1])//如果没有颜色就加一，如果与之前相等那么也是加一                i++;            res[sum[i]]++;        }        for(int i=0; i<maxn; i++)if(res[i])printf("%d %d\n",i,res[i]);        printf("\n");    }    return 0;}