【BZOJ 2809】【APIO 2012】dispatching

昨天晚上zyf神犇问我的题，虽然我太弱参加不了APIO但也做一做吧。

用小数据拍了无数次总是查不出错来，交上去就WA，后来用国内数据测发现是主席树上区间相减的值没有用long long存，小数据真是没用啊QuQ

手残不加long long毁一生

交上去还是WA，最后发现没有把windows下的I64d改成linux下的lld==

因为这个贡献了两次Wrong Answer，，，好像第一次还爆内存了？比赛时这么手残就得滚粗了QAQ

#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
using namespace std;
const int N = 100003;
int getint() {
int k = 0, fh = 1; char c = getchar();
for(; c < '0' || c > '9'; c = getchar())
if (c == '-') fh = -1;
for(; c >= '0' && c <= '9'; c = getchar())
k = k * 10 + c - '0';
return k * fh;
}
struct node1 {
int b, c, l, id;
} Q
;
struct node2 {
int l, r, s;
long long sc;
}T[N * 30];
int S
, n, m, point
, nxt
, to
, cnt = 0, rt = 0, L
, R
, H
;
int root
, CL
;
void ins(int x, int y) {nxt[++cnt] = point[x]; to[cnt] = y; point[x] = cnt;}
void dfs(int x) {
S[++cnt] = x;
L[x] = cnt;
for(int i = point[x]; i; i = nxt[i]) dfs(to[i]);
R[x] = cnt;
}
bool cmp(node1 X, node1 Y) {return X.c < Y.c;}
void update(int l, int r, int &pos, int key) {
T[++cnt] = T[pos]; pos = cnt; ++T[pos].s; T[pos].sc += Q[key].c;
if (l == r) return;
int mid = (l + r) >> 1;
if (key <= mid) update(l, mid, T[pos].l, key);
else update(mid + 1, r, T[pos].r, key);
}
long long Query(int l, int r, int x, int y, int key) {
if (l == r) {
if (T[y].sc - T[x].sc <= key) return T[y].s - T[x].s;
else return key / Q[l].c;
}
int mid = (l + r) >> 1; long long su = T[T[y].l].sc - T[T[x].l].sc;
if (su >= key) return Query(l, mid, T[x].l, T[y].l, key);
else return T[T[y].l].s - T[T[x].l].s + Query(mid + 1, r, T[x].r, T[y].r, key - su);
}
int main() {
read(n); read(m);
for(int i = 1; i <= n; ++i)
read(Q[i].b), read(Q[i].c), read(Q[i].l), Q[i].id = i, CL[i] = Q[i].l;
for(int i = 1; i <= n; ++i)
if (Q[i].b == 0) rt = i;
else ins(Q[i].b, i);
cnt = 0;
dfs(rt);
sort(Q + 1, Q + n + 1, cmp);
for(int i = 1; i <= n; ++i)
H[Q[i].id] = i;
cnt = 0;
//	for(int i = 1; i <= n; ++i) printf("%d %d\n", L[i], R[i]);
//	for(int i = 1; i <= n; ++i)	printf("!! %d\n", H[i]);
//	for(int i = 1; i <= n; ++i)	printf("!! %d\n", Q[H[i]].c);
for(int i = 1; i <= n; ++i) {
root[i] = root[i - 1];
update(1, n, root[i], H[S[i]]);
//		printf("HHH %d\n", S[i]);
//		printf("!!! %d\n", H[S[i]]);
}
//	for(int i = 1; i <= n ;++i) printf("%I64d ", T[root[i]].sc); puts("");
long long ans = 0;
for(int i = 1; i <= n; ++i)
ans = max(ans, CL[i] * Query(1, n, root[L[i] - 1], root[R[i]], m));
printf("%lld\n", ans);
//	for(int i = 1; i <= n; ++i) printf("%d %d\n", L[i], R[i]);
return 0;
}

没有删掉愚蠢的调试信息，，，