hdu4553（神TM烦的线段树维护连续区间ORZ）

/*
translation:
中文题意，不表
solution:
线段树维护连续区间
此题还是按照基本的线段树维护连续区间的方式来，但是有一点较为复杂，就是要同时维护
DS和NS两个线段树。所以要清楚二者之间的关系，NS能够影响到DS的树，而DS不能影响到
NS的区间。注意合并区间过程中极易出错。
note:
#: 之前WA了几次都是因为m=(s[o]+e[o])>>1和m=(l+r)>>1傻傻分不清
date:
2016.11.27
*/
#include <iostream>
#include <cstdio>

using namespace std;
const int maxn = 100000 +5;

int n, q;
int s[maxn*4][2], e[maxn*4][2];
int preLen[maxn*4][2], sufLen[maxn*4][2], maxLen[maxn*4][2];
int cover[maxn*4][2];

void build(int t, int o, int l, int r)
{
s[o][t] = l;
e[o][t] = r;
preLen[o][t] = maxLen[o][t] = sufLen[o][t] = r - l + 1;
cover[o][t] = -1;
int m = (l + r) >> 1;

if(l == r)	return;
build(t, o << 1, l, m);
build(t, o << 1 | 1, m+1, r);
}

void push_down(int t, int o, int num)
{
int lc = o << 1;
int rc = o << 1 | 1;

if(cover[o][t] != -1){
if(cover[o][t] == 0){
preLen[lc][t] = maxLen[lc][t] = sufLen[lc][t] = 0;
preLen[rc][t] = maxLen[rc][t] = sufLen[rc][t] = 0;
cover[lc][t] = cover[rc][t] = 0;
cover[o][t] = -1;
}
else{
preLen[lc][t] = maxLen[lc][t] = sufLen[lc][t] = num - (num >> 1);
preLen[rc][t] = maxLen[rc][t] = sufLen[rc][t] = num >> 1;
cover[lc][t] = cover[rc][t] = 1;
cover[o][t] = -1;
}
}
}

void push_up(int t, int o, int num)
{
int lc = o << 1, rc = o << 1 | 1;
preLen[o][t] = preLen[lc][t];
sufLen[o][t] = sufLen[rc][t];

if(preLen[lc][t] == num - (num >> 1))	preLen[o][t] += preLen[rc][t];
if(sufLen[rc][t] == num >> 1)	sufLen[o][t] += sufLen[lc][t];

maxLen[o][t] = max(maxLen[lc][t], maxLen[rc][t]);
maxLen[o][t] = max(maxLen[o][t], sufLen[lc][t] + preLen[rc][t]);
}

void update(int t, int l, int r, int c, int o)
{
if(s[o][t] == l && e[o][t] == r){
cover[o][t] = c;
if(c)	preLen[o][t] = maxLen[o][t] = sufLen[o][t] = r - l + 1;
else	preLen[o][t] = maxLen[o][t] = sufLen[o][t] = 0;
return;
}
if(s[o][t] == e[o][t])	return;

push_down(t, o, e[o][t] - s[o][t] + 1);
int m = (s[o][t] + e[o][t]) >> 1;
if(r <= m)	update(t, l, r, c, o << 1);
else if(l > m)	update(t, l, r, c, o << 1 | 1);
else{
update(t, l, m, c, o << 1);
update(t, m + 1, r, c, o << 1 | 1);
}
push_up(t, o, e[o][t] - s[o][t] + 1);
}

int query(int t, int len, int o, int l, int r)
{
if(l == r)	return l;
push_down(t, o, e[o][t] - s[o][t] + 1);

int m = (s[o][t] + e[o][t]) >> 1;
int lc = o << 1, rc = o << 1 | 1;

if(maxLen[lc][t] >= len)	return query(t, len, lc, 1, m);
else if(sufLen[lc][t] + preLen[rc][t] >= len)	return m - sufLen[lc][t] + 1;
else	return query(t, len, rc, m + 1, r);
}

int main()
{
//freopen("in.txt", "r", stdin);
int T, kase = 0;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &q);
build(0, 1, 1, n);	//ds
build(1, 1, 1, n);	//ns

char op[10];
int x, y;
printf("Case %d:\n", ++kase);

while(q--){
scanf("%s", op);
if(op[0] == 'D'){
scanf("%d", &x);
if(maxLen[1][0] >= x){
int ans = query(0, x, 1, 1, n);
printf("%d,let's fly\n", ans);
update(0, ans, ans + x - 1, 0, 1);
}
else	printf("fly with yourself\n");
}
else if(op[0] == 'N'){
scanf("%d", &x);
if(maxLen[1][0] >= x){
int ans = query(0, x, 1, 1, n);
//printf("#:update: %d %d\n", ans, ans + x - 1);
printf("%d,don't put my gezi\n", ans);
update(0, ans, ans + x - 1, 0, 1);
update(1, ans, ans + x - 1, 0, 1);
}
else{
if(maxLen[1][1] >= x){
int ans = query(1, x, 1, 1, n);
//printf("#:update: %d %d\n", ans, ans + x - 1);
printf("%d,don't put my gezi\n", ans);
update(0, ans, ans + x - 1, 0, 1);
update(1, ans, ans + x - 1, 0, 1);
}
else printf("wait for me\n");
}
}
else{
scanf("%d%d", &x, &y);
printf("I am the hope of chinese chengxuyuan!!\n");
update(0, x, y, 1, 1);
update(1, x, y, 1, 1);
}
}
}
return 0;
}