POJ 2726 + 1828

/*输入n个二元组(xi,yi)，统计有多少个二元组(xk,yk)满足：*对于任意xi < xk的二元组，yi > yk；*对于任意yi < yk的二元组, xi > xk*( n<= 10000，ai<= 10000)*排序，然后统计。*/#include <iostream>#include <algorithm>using namespace std;const int maxn = 10001;struct point{int d;int c;};point arr[maxn];int N, nCount;void funcInput();void funcSolve();void funOutput();bool cmp(const point &a, const point &b);int main(){while (1 == scanf("%d", &N), N){funcInput();funcSolve();funOutput();}return 0;}void funcInput(){for(int i = 1; i <= N; i++){scanf("%d %d", &arr[i].d, &arr[i].c);}}void funcSolve(){int i;/*按distance从小到大排列*/sort(&arr[1], &arr[N+1], cmp);int nTemp = arr[1].c;nCount = 1;for(i = 2; i <= N; i++){if(arr[i].c < nTemp){nCount ++;nTemp = arr[i].c;}}}void funOutput(){printf("%d/n", nCount);}bool cmp(const point &a, const point &b){if(a.d == b.d){return a.c < b.c;}else{return a.d < b.d;}}

 

poj 1828

/*输入n个二元组(xi,yi)，统计有多少个二元组(xk,yk)满足：*对于任意xi > xk的二元组，yi > yk；*( n<= 10000，ai<= 10000)*排序，然后统计。*/#include <iostream>#include <algorithm>using namespace std;const int maxn = 50001;struct point{int d;int c;};point arr[maxn];int N, nCount;void funcInput();void funcSolve();void funOutput();bool cmp(const point &a, const point &b);int main(){while (1 == scanf("%d", &N), N){funcInput();funcSolve();funOutput();}return 0;}void funcInput(){for(int i = 1; i <= N; i++){scanf("%d %d", &arr[i].d, &arr[i].c);}}void funcSolve(){int i;/*按distance从小到大排列*/sort(&arr[1], &arr[N+1], cmp);int nTemp = arr[1].c;nCount = 1;for(i = 2; i <= N; i++){if(arr[i].c > nTemp){nCount ++;nTemp = arr[i].c;}}}void funOutput(){printf("%d/n", nCount);}bool cmp(const point &a, const point &b){if(a.d == b.d){return a.c > b.c;}else{return a.d > b.d;}}