POJ 1151 && 1389 离散化求矩形面积的并

两个题目都是离散化的基本模型。解题思路：

1、首先分离出所有的横坐标和纵坐标分别按升序存入数组X[ ]和Y[ ]中.
2、 设数组XY[ ][ ].对于每个矩形(x1,y1)(x2,y2)确定i1,i2,j1,j2,使得,X[i1]>=x1,X[i2]<=x2,Y[i1]>y1,Y[i2]>=y2令XY[ i ][ j ] = 1 (i从[i1,i2)，j从[j1,j2))
3、统计面积：area+=XY[i][j] *(X[i]-X[i-1])*(Y[i] – Y[i-1])

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 1000
#define INF 0x7FFFFFFF
#define REP(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define L(x) x<<1
#define R(x) x<<1|1
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;
int n;
struct node {
int x1,y1,x2,y2;
} p[MAX];

bool vis[2*MAX][2*MAX];
int x[2*MAX],y[2*MAX];
int main() {

int i=1,cnt = 0,n;
int casee = 1;
while(1) {
i = 1;
scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
if(p[i].x1 == -1) break;
i++;
while(scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2)) {
if(p[i].x1 == -1) break;
i++;
}
n = i;
memset(vis,0,sizeof(vis));
cnt = 0;
for(i=1; i<=n; i++) {
x[cnt] = p[i].x1;
y[cnt] = p[i].y1;
cnt++;
x[cnt] = p[i].x2;
y[cnt] = p[i].y2;
cnt++;
}
sort(x,x+cnt);
sort(y,y+cnt);
for(i=1; i<=n ; i++) {
int a,b,c,d;
for(a=0; a<cnt; a++) {
if(x[a] == p[i].x1) break;
}
for(b=0; b<cnt; b++) {
if(y[b] == p[i].y1) break;
}
for(c=0; c<cnt; c++) {
if(x[c] == p[i].x2) break;
}
for(d=0; d<cnt; d++) {
if(y[d] == p[i].y2) break;
}
for(int j=a; j<c; j++) {
for(int k=b; k<d; k++) {
vis[j][k] = 1;
}
}
}
int s = 0;
for(i=0; i<cnt; i++) {
for(int j=0; j<cnt; j++) {
if(vis[i][j]) s += (x[i+1] -x[i])*(y[j+1] -y[j]);
}
}
printf("%d\n",s);
}
return 0;
}