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HDU 3333 & 3874 (线段树+离线询问)

2013-08-02 22:46 337 查看
两个题目都是求区间之内,不重复的数字之和,3333需要离散化处理.................

调试了一下午........说多了都是泪...........

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <climits>//形如INT_MAX一类的
#define MAX 51111
#define INF 0x7FFFFFFF
#define L(x) x<<1
#define R(x) x<<1|1
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;

inline void RD(int &ret) {
char c;
do {
c = getchar();
} while(c < '0' || c > '9') ;
ret = c - '0';
while((c=getchar()) >= '0' && c <= '9')
ret = ret * 10 + ( c - '0' );
}

inline void OT(int a) {
if(a >= 10)OT(a / 10) ;
putchar(a % 10 + '0') ;
}

struct node {
int l,r,mid,cover;
__int64 sum;
} tree[MAX*4];

int n,m;
int a[MAX],va[MAX],pos[MAX],tmp[MAX];
__int64 ans[111111 * 2];

struct Node {
int l,r;
int id;
} qes[111111 * 2];

void init() {
memset(va,0,sizeof(va));
}
int search(int l,int r,int x) {
int mid;
while(l <= r) {
mid = (l+r) >> 1;
if(pos[mid] == x) return mid;
else if(pos[mid] > x) r = mid -1;
else l = mid + 1;
}
}

void up(int num) {
tree[num].sum = tree[L(num)].sum + tree[R(num)].sum;
}

void build(int l,int r,int num) {
tree[num].l = l;
tree[num].r = r;
tree[num].mid = (l+r) >> 1;
tree[num].cover = 0;
tree[num].sum = 0;
if(l == r) {
tree[num].sum = va[l];
return ;
}
build(l,tree[num].mid,L(num));
build(tree[num].mid + 1,r,R(num));
up(num);
}

void update(int l,int x,int color) {
if(tree[x].l == tree[x].r) {
tree[x].sum += color;
return ;
}
if(l > tree[x].mid) update(l,R(x),color);
else update(l,L(x),color);
up(x);
}

__int64 query(int l,int r,int num) {
if(l == tree[num].l && tree[num].r == r) {
return tree[num].sum;
}
if(r <= tree[num].mid) {
return query(l,r,L(num));
} else if(l > tree[num].mid) {
return query(l,r,R(num));
} else {
return query(l,tree[num].mid,L(num)) + query(tree[num].mid+1,r,R(num));
}
}

bool cmp(const Node &a,const Node &b) {
if(a.r == b.r) return a.l < b.l;
return a.r < b.r;
}

bool cmp2(const int &a, const int &b) {
return a < b;
}
int main() {
int T;
cin >> T;
while(T --) {
init();
RD(n);
int t = 1;
for(int i=1; i<=n; i++) {
RD(a[i]);
tmp[i] = a[i];
}
sort(tmp+1,tmp+1+n,cmp2);
pos[1] = tmp[1];
for(int i=2; i<=n; i++) {
if(tmp[i] != tmp[i-1]) {
pos[++t] = tmp[i];
}
}
build(1,n,1);
RD(m);
for(int i=1; i<=m; i++) {
RD(qes[i].l);
RD(qes[i].r);
qes[i].id = i;
}
sort(qes+1,qes+1+m,cmp);
int order = 1;
for(int i=1; i<=m; i++) {
while(qes[i].r >= order) {
int id = search(1,t,a[order]);
int ps = va[id];
if( ps != 0) update(ps,1,-a[order]);
va[id] = order;
update(va[id],1,a[order]);
order ++;
}
ans[qes[i].id] = query(qes[i].l,qes[i].r,1);
}
for(int i=1; i<=m; i++) {
printf("%I64d\n",ans[i]);
}
}
return 0;
}
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