## Sunday, December 27, 2015

### poj 2481 Cows

http://poj.org/problem?id=2481
Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good.

Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].

But some cows are strong and some are weak. Given two cows: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj.

For each cow, how many cows are stronger than her? Farmer John needs your help!
Input
The input contains multiple test cases.
For each test case, the first line is an integer N (1 <= N <= 105), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 105) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.

The end of the input contains a single 0.
Output
For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cowi
Sample Input
```3
1 2
0 3
3 4
0
```
Sample Output
`1 0 0`

 `009` `   ``class` `TNode ``implements` `Comparable{`
 `010` `    ``int` `s;``//左端点`
 `011` `    ``int` `e;``//右端点`
 `012` `    ``int` `label;``//牛的序号`
 `013` `    ``public` `int` `compareTo(Object o) { `
 `014` `       ``int` `v=((TNode)o).e;`
 `015` `       ``if``(``this``.e!=v) ``//按右端点降序排列`
 `016` `         ``return` `v-``this``.e;  `
 `017` `       ``return` `this``.s-((TNode)o).s;``//右端点相等,按左端点升序排序`
 `018` `    ``}`
 `019`
 `020` `    ``public` `String toString(){`
 `021` `        ``return` `(``"["``+s+``","``+e+``"]"``);`
 `022` `    ``}`
 `023` `  ``}`
 `024`
 `025` ` ``public` `class` `Main{`
 `026` `   ``static`  `int` `N=``100015``;`
 `027` `   ``TNode[] cow;`
 `028` `   ``int` `cal[];``//树状数组`
 `029` `   ``int` `res[];`
 `030` `   ``int` `maxn;`
 `031`
 `032` `  ``public` `Main(){`
 `033` `    `
 `034` `     ``}`
 `035` `       `
 `036` `  `
 `037`
 `038` `  ``private` `int` `lowbit(``int` `t){``//计算cal[t]展开的项数  `
 `039` `   ``return` `t&(-t);  `
 `040` `  ``}  `
 `041`
 `042` `  ``private` `int` `Sum(``int` `k){ ``//求前k项的和.  `
 `043` `   ``int` `sum=``0``;   `
 `044` `    ``while``(k>``0``){   `
 `045` `       ``sum+=cal[k];   `
 `046` `       ``k=k-lowbit(k);   `
 `047` `    ``}       `
 `048` `    ``return` `sum;   `
 `049` `  ``}   `
 `050`
 `051` `  ``private` `void` `update(``int` `i,``int` `x){    ``//增加某个元素的大小,给某个节点 i 加上 x  `
 `052` `      ``while``(i<=maxn){   `
 `053` `        ``cal[i]=cal[i]+x; ``//更新父节点`
 `054` `        ``i=i+lowbit(i);   `
 `055` `     ``}   `
 `056` `    ``}   `
 `061` `  ``public` `static` `void`  `main(String args[]) ``throws` `IOException{`
 `062` `        ``Main ma=``new` `Main();`
 `063` `        ``ma.go();`
 `064` `   ``}`
 `066` `    ``public` `void` `go() ``throws` `IOException{`
 `067` `     `
 `068` `      ``StreamTokenizer st = ``new` `StreamTokenizer(``new` `BufferedReader(     `
 `069` `      ``new` `InputStreamReader(System.in)));     `
 `070` `      ``PrintWriter out = ``new` `PrintWriter(``new` `OutputStreamWriter(System.out));     `
 `071`
 `072` `      ``while``(``true``) {`
 `073` `       ``st.nextToken();     `
 `074` `      ``int` `n= (``int``) st.nval;    ``//牛的个数 `
 `075` `      ``if``(n==``0``) ``break``;`
 `076`
 `077` `     ``cow=``new` `TNode[N];`
 `078` `     ``cal=``new` `int``[N];`
 `079` `     ``res=``new` `int``[N];`
 `080` `     ``for``(``int` `i=``0``;i< N;i++){`
 `081` `       ``cow[i]=``new` `TNode();`
 `082` `      ``// cal[i]=0;`
 `083` `      ``}`
 `084` `      ``for``(``int` `i=``0``;i< n;i++) {``//读入牛的吃草区间`
 `085` `            ``st.nextToken();    `
 `086` `           ``cow[i].s=(``int``) st.nval;`
 `087` `             ``st.nextToken();    `
 `088` `           ``cow[i].e=(``int``) st.nval;`
 `089` `           ``cow[i].s++;`
 `090` `           ``cow[i].e++;`
 `091` `           ``cow[i].label=i;``//牛的原始序号`
 `092` `           ``if``(cow[i].e>maxn) maxn=cow[i].e;``//最大右端点`
 `093` `      ``}`
 `094` `       `
 `095` `        ``Arrays.sort(cow);``//排序`
 `096`
 `097` `     ``// for(int i=0;i< n;i++)`
 `098` `       ``// System.out.println(cow[i]);`
 `099` ` `
 `100` `      ``for``(``int` `i=``0``;i< n;i++) {`
 `101` `           ``if``(i!=``0``&&cow[i].s==cow[i-``1``].s&&cow[i].e==cow[i-``1``].e)``//两头牛有相同的吃草区间`
 `102` `                ``res[cow[i].label]=res[cow[i-``1``].label];``//它们有相同的答案`
 `103` `            ``else` `res[cow[i].label]=Sum(cow[i].s);``//统计比cow[i].label这头牛强的牛的数目`
 `104` `            ``update(cow[i].s,``1``);``//更新`
 `105` `        ``}`
 `106` `        ``System.out.printf(``"%d"``,res[``0``]);`
 `107` `        ``for``(``int` `i=``1``;i< n;i++) System.out.printf(``" %d"``,res[i]);`
 `108` `        ``System.out.println();`
 `109` `    ``}`
 `110` `  ``}`