ZOJ 2301 Color the Ball 线段树 区间染色

做法：这题比ural那题简单，因为是球，所以离散化时记录区间边间点和中间的两个点就好了。

//列SUM组的方法也只适用于，空出的的格子可以不用考虑的情况啊#include<cstdio>#include<cstring>#include<algorithm>#define left l,m,x<<1#define right m+1,r,x<<1|1using namespace std;const int LMT=12005;const int eps=1<<31-1;int cov[LMT<<2],myx[LMT],leng,cnt;bool seg[LMT];struct line{int a,b;char c;}e[LMT];inline int max(int a,int b){return a>b?a:b;}void init(void){cov[1]=1;memset(seg,1,sizeof(seg));cnt=0;leng=0;}void cut(int x){if(cov[x]!=-1){cov[x<<1]=cov[x<<1|1]=cov[x];cov[x]=-1;}}void update(int op,int L,int R,int l,int r,int x){if(L<=l&&r<=R){cov[x]=op;return;}cut(x);int m=(l+r)>>1;if(L<=m)update(op,L,R,left);if(R>m)update(op,L,R,right);}void query(int l,int r,int x){if(cov[x]!=-1){int i;for(i=l;i<=r;i++)seg[i]=cov[x];return;}if(l==r)return;cut(x);int m=(l+r)>>1;query(left);query(right);}int main(void){int i,j,ansl,ansr,n,t;char c;while(~scanf("%d",&n)){init();for(i=0;i<n;i++){scanf("%d%d",&e[i].a,&e[i].b);myx[cnt++]=e[i].a;myx[cnt++]=e[i].b;c=getchar();while(c!='w'&&c!='b')c=getchar();e[i].c=c;}sort(myx,myx+cnt);t=1;for(i=1;i<cnt;i++)if(myx[i]!=myx[i-1])myx[t++]=myx[i];cnt=t;        for(i=1;i<t;i++){if(myx[i]-myx[i-1]>1)myx[cnt++]=myx[i-1]+1;if(myx[i]-myx[i-1]>2)myx[cnt++]=myx[i]-1;}sort(myx,myx+cnt);myx[cnt++]=eps;for(i=0;i<n;i++){e[i].a=find(myx,myx+cnt,e[i].a)-myx;e[i].b=find(myx,myx+cnt,e[i].b)-myx;update(e[i].c=='b',e[i].a,e[i].b,0,cnt-1,1);}query(0,cnt-1,1);for(i=0;i<cnt;)if(seg[i]==0){j=i;while(j<cnt&&seg[i]==seg[j])j++;if(myx[j-1]-myx[i]+1>leng){leng=myx[j-1]-myx[i]+1;ansl=myx[i];ansr=myx[j-1];}i=j;}else i++;if(leng)printf("%d %d\n",ansl,ansr);else puts("0");}return 0;}