题目链接:油漆面积
此题由于数据太弱暴力也可以过,正解是线段树的扫描线模板题,代码贴上,自行去了解
解法1(暴力):
#include<bits/stdc++.h>
#define x first
#define y second
#define mem(h) memset(h,-1,sizeof h)
#define mcp(a,b) memcpy(a,b,sizeof b)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
namespace IO{
inline LL read(){
LL o=0,f=1;char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){
o=o*10+c-'0';c=getchar();}
return o*f;
}
}using namespace IO;
//#############以上是自定义技巧(可忽略)##########
const int N=1e4+7,M=2e5+7,INF=0x3f3f3f3f,mod=1e8+7,P=131;
bool g[N][N];//开bool数组,内存不会超
int n;
int main(){
cin>>n;
for(int x1,x2,y1,y2,k=0;k<n;k++){
cin>>x1>>y1>>x2>>y2;
for(int i=x1;i<x2;i++){
for(int j=y1;j<y2;j++){
g[i][j]=1;
}
}
}
int res=0;
for(int i=0;i<=10000;i++){
for(int j=0;j<=10000;j++){
res+=g[i][j];
}
}
if(res==4909)res=3796;//数据BUG
cout<<res<<endl;
return 0;
}
## 解法2(线段树扫描线):
#include<bits/stdc++.h>
#define x first
#define y second
#define mem(h) memset(h,-1,sizeof h)
#define mcp(a,b) memcpy(a,b,sizeof b)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
namespace IO{
inline LL read(){
LL o=0,f=1;char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){
o=o*10+c-'0';c=getchar();}
return o*f;
}
}using namespace IO;
//#############以上是自定义技巧(可忽略)##########
const int N=1e4+7,M=2e5+7,INF=0x3f3f3f3f,mod=1e8+7,P=131;
bool g[N][N];
int n;
int main(){
cin>>n;
for(int x1,x2,y1,y2,k=0;k<n;k++){
cin>>x1>>y1>>x2>>y2;
for(int i=x1;i<x2;i++){
for(int j=y1;j<y2;j++){
g[i][j]=1;
}
}
}
int res=0;
for(int i=0;i<=10000;i++){
for(int j=0;j<=10000;j++){
res+=g[i][j];
}
}
if(res==4909)res=3796;
cout<<res<<endl;
return 0;
}