POJ_2481（树状数组）

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本题的关键在于要考虑到两个区间完全相同的情况，其他的不用考虑太多。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;const int maxn = 100000 + 300;int c[maxn];int ans[maxn];struct cow{int s;int e;int in;bool operator < (const cow& z) const{if(z.s == s) return z.e < e;else return z.s > s;}}co[maxn], pri;int lowbit(int x){return x & (-x);}int sum(int x){int t = 0;while(x > 0){t += c[x];x -= lowbit(x);}return t;} int update(int x){while(x < maxn){c[x]++;x += lowbit(x);}return 0;}int main(){int tot;while(scanf("%d", &tot) && tot){memset(c, 0, sizeof(c));memset(co, 0, sizeof(co));memset(ans, 0, sizeof(ans));for(int i = 0; i < tot; ++i){scanf("%d%d", &co[i].s, &co[i].e);co[i].s++; co[i].e++;co[i].in = i;}sort(co, co+tot);pri.s = co[0].s; pri.e = co[0].e; pri.in = co[0].in;co[0].e++;ans[co[0].in] = 0 - sum(co[0].e);update(co[0].e);for(int i = 1; i < tot; ++i){ int s = co[i].e;s++;if(pri.e == co[i].e && pri.s == co[i].s){ans[co[i].in] = ans[co[i-1].in];}else{ans[co[i].in] = i - sum(s-1);}pri.e = co[i].e; pri.s = co[i].s; pri.in = co[i].in;update(s);}int i;for(i = 0; i < tot - 1; ++i){printf("%d ", ans[i]);} printf("%d\n", ans[i]);}return 0;}

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