hdu3872

source: http://acm.hdu.edu.cn/showproblem.php?pid=3872

title:    Dragon Ball

解法：线段树

这题的题解不太好写。需要利用的一个特点是：

设当前扫描到点i，vMax[u]表示[u, i]之间的能量的最大值，则vMax具有

不增性。

 

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int N = 100005;
const ll INF = 0X7FFFFFFFFFFFFFFFLL;
struct seg{
	int l, r;
	int dv;
	ll best;
	int mid(){
		return (l + r) >> 1;
	}
}segs[N<<2];

int v[N];
int last[N], r[N], vMax[N];
int n;
template <typename T>
void getNum(T& a){
	a = 0;
	char ch;
	while(true){
		ch = getchar();
		if(ch >= '0' && ch <= '9') break;
	}
	a = ch - '0';
	while(true){
		ch = getchar();
		if(ch < '0' || ch > '9') break;
		a = a * 10 + ch - '0';
	}
}

inline int ls(int id){
	return (id << 1) + 1;
}

inline int rs(int id){
	return (id << 1) + 2;
}
void build(int id, int l, int r){
	segs[id].l = l; segs[id].r = r;
	segs[id].dv = 0; segs[id].best = INF;
	if(l < r){
		int mid = segs[id].mid();
		build(ls(id), l, mid);
		build(rs(id), mid+1, r);
	}
}
bool input(){
	getNum(n);
	int i, tMax, t;
	tMax = 0;
	for(i = 1; i <= n; i++){
		getNum(t);
		if(t > tMax){
			do{
				last[++tMax] = -1;
			}while(tMax < t);
		}
		if(last[t] == -1){
			r[i] = 1;
			last[t] = i;
		}else{
			r[i] = last[t] + 1;
			last[t] = i;
		}
	}
	for(i = 1; i <= n; i++){
		getNum(v[i]);
	}
	return true;
}
inline void up(int id, int dv){
	segs[id].best += dv;
	segs[id].dv += dv;
}
void down(int id){
	if(segs[id].dv != 0){
		up(ls(id), segs[id].dv);
		up(rs(id), segs[id].dv);
		segs[id].dv = 0;
	}
}
void update(int id){
	segs[id].best = INF;
	if(segs[id].best > segs[ls(id)].best){
		segs[id].best = segs[ls(id)].best;
	}
	if(segs[id].best > segs[rs(id)].best){
		segs[id].best = segs[rs(id)].best;
	}
}
void ins(int id, int pos, ll v){
	if(segs[id].l == segs[id].r){
		segs[id].best = v;
		return ;
	}
	down(id);
	if(pos <= segs[id].mid()){
		ins(ls(id), pos, v);
	}else{
		ins(rs(id), pos, v);
	}
	update(id);
}
void ins(int id, int l, int r, ll dv){
	if(l <= segs[id].l && segs[id].r <= r){
		up(id, dv);
		return;
	}
	if(segs[id].l == segs[id].r) return;
	down(id);
	int mid = segs[id].mid();
	if(l <= mid){
		ins(ls(id), l, r, dv);
	}
	if(r > mid){
		ins(rs(id), l, r, dv);
	}
	update(id);
}
ll qry(int id, int l, int r){
	if(l <= segs[id].l && segs[id].r <= r){
		return segs[id].best;
	}
	if(segs[id].l == segs[id].r) return INF;
	down(id);
	int mid = segs[id].mid();
	ll ans = INF;
	if(l <= mid){
		ans = min(ans, qry(ls(id), l, r));
	}
	if(r > mid){
		ans = min(ans, qry(rs(id), l, r));
	}
	update(id);
	return ans;
}
void solve(){
	build(0, 1, n);
	int i, u;
	ll lv;
	lv = 0;
	for(i = 1; i <= n; i++){
		ins(0, i, lv + v[i]);
		last[i] = i;
		vMax[i] = v[i];
		for(u = i; u >= 1 && vMax[u] <= v[i]; ){
			if(vMax[u] != v[i]){
				ins(0, last[u], u, v[i] - vMax[u]);
			}
			last[i] = last[u];
			u = last[u] - 1;
		}
		lv = qry(0, r[i], i);
	}
	printf("%I64d\n", lv);
}
int main() {
	int t;
	getNum(t);
	while(t--){
		input();
		solve();
	}
	return 0;
}