本文主要是介绍poj 2398 Toy Storage（计算几何：叉积），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
基本上和poj 2318一模一样。。。
改下输出就可以了
代码如下：
/* ***********************************************
Author :yinhua
Created Time :2014年12月01日 星期一 19时25分15秒
File Name :poj2398.cpp
************************************************ */#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAXN 100010
#define LL long long
using namespace std;struct Point { int x, y; Point() { } Point(int _x, int _y) { x = _x;y = _y; } Point operator -(const Point &b) const { return Point(x-b.x, y-b.y); } int operator *(const Point &b) const { return x*b.x+y*b.y; } int operator ^(const Point &b) const { return x*b.y-y*b.x; }
}; struct Line { Point s, e; Line() { } Line(Point _s, Point _e) { s = _s; e = _e; }
}; bool cmp(Line a, Line b) {return (a.s).x < (b.s).x;
}int xmult(Point p0, Point p1, Point p2) {//p0点与线段p1 p2的叉积 return (p1-p0)^(p2-p0);
} Line line[MAXN];
int res[MAXN];
int ans[MAXN]; int main(void) { int m, n, x1, y1, x2, y2; bool first = true; while(~scanf("%d", &n) && n) { scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2); int ui, li; for(int i=0; i<n; ++i) { scanf("%d%d", &ui, &li); line[i] = Line(Point(ui, y1), Point(li, y2)); } sort(line, line+n, cmp);line[n] = Line(Point(x2, y1), Point(x2, y2)); int x, y; Point p; memset(ans, 0, sizeof(ans)); for(int i=1; i<=m; ++i) {scanf("%d%d", &x, &y); p = Point(x, y); int l = 0, r = n; int tmp; while(l <= r) { int mid = (l+r)>>1; if(xmult(p, line[mid].s, line[mid].e) < 0) {//点在当前线的左侧 tmp = mid; r = mid-1; } else l = mid+1;//点在当前线的右侧 } ans[tmp]++; } memset(res, 0, sizeof(res));for(int i=0; i<=n; ++i) {if(ans[i]) ++res[ans[i]];}bool ok = true;for(int i=1; i<=m; ++i) {if(res[i]) {if(ok) {ok = false;puts("Box");}printf("%d: %d\n", i, res[i]);}}} return 0;
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