BZOJ1218(NOI2003)[激光炸弹]--容斥

【链接】
bzoj1218

【解题报告】

容斥裸题吧。。

#include<cstdio>#include<algorithm>using namespace std;const int maxn=5005;int n,m,ans,E,K,f[maxn][maxn];inline char nc(){    static char buf[100000],*l,*r;    if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);    if (l==r) return EOF; return *l++;}inline int Read(){    int res=0,f=1; char ch=nc(),cc=ch;    while (ch<'0'||ch>'9') cc=ch,ch=nc();    if (cc=='-') f=-1;    while (ch>='0'&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}int main(){    freopen("1218.in","r",stdin);    freopen("1218.out","w",stdout);    n=m=ans=0; E=Read(); K=Read();    for (int i=1,x,y; i<=E; i++) x=Read()+1,y=Read()+1,n=max(n,x),m=max(m,y),f[x][y]=Read();    if (n<K) n=K; if (m<K) m=K;    for (int i=1; i<=n; i++)     for (int j=1; j<=m; j++)      f[i][j]+=f[i-1][j]+f[i][j-1]-f[i-1][j-1];    for (int i=0; i<=n-K; i++)     for (int j=0; j<=m-K; j++)      ans=max(ans,f[i+K][j+K]-f[i][j+K]-f[i+K][j]+f[i][j]);    printf("%d",ans);    return 0;}