HDU 1566 Color the ball

Color the ballTime Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18042    Accepted Submission(s): 8998Problem DescriptionN个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？Input每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。当N = 0，输入结束。Output每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。Sample Input31 12 23 331 11 21 30Sample Output1 1 13 2 1Author8600SourceHDU 2006-12 Programming Contest

解法①

非常巧妙的解法…
a[i]表示 以i作为起点 对i-n的气球全部上色的次数
那对(start,end)
++a[start]
--a[end+1] 抵消掉end+1 - n 的部分
那问题就转化为求a的前缀和

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#include<iostream>#include<stdlib.h>#include<stdio.h>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<time.h>#include<math.h>#include<list>#include<cstring>#include<fstream>//#include<memory.h>using namespace std;#define ll long long#define ull unsigned long long#define pii pair<int,int>#define INF 1000000007#define pll pair<ll,ll>#define pid pair<int,double>const int N = 100000;int a[N];int main(){    //freopen("/home/lu/文档/r.txt","r",stdin);    //freopen("/home/lu/文档/w.txt","w",stdout);    int n,s,e;    while(scanf("%d",&n),n){        fill(a,a+n+1,0);        for(int i=0;i<n;++i){            scanf("%d%d",&s,&e);            ++a[s],--a[e+1];        }        printf("%d",a[1]);        for(int i=2;i<=n;++i){            a[i]+=a[i-1];            printf(" %d",a[i]);        }        putchar('\n');    }    return 0;}

解法②

线段树区间更新

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#include<iostream>#include<stdlib.h>#include<stdio.h>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<time.h>#include<math.h>#include<list>#include<cstring>#include<fstream>//#include<memory.h>using namespace std;#define ll long long#define ull unsigned long long#define pii pair<int,int>#define INF 1000000007#define pll pair<ll,ll>#define pid pair<int,double>const int N = 100000+5;struct{    int l,r,n;}rmq[4*N];void init_rmq(int l,int r,int k){    rmq[k]={l,r,0};    if(l!=r){        int mid = (l + r)/2;        init_rmq(l,mid,2*k);        init_rmq(mid + 1,r,2*k + 1);    }}void update(int s,int e,int k,int x){    if(s==rmq[k].l&&e==rmq[k].r){        rmq[k].n+=x;        return;    }    int mid = (rmq[k].l + rmq[k].r)/2;    if(s<=mid){//左子数[l,mid]        update(s,min(e,mid),2*k,x);    }    if(mid+1<=e){//右子树[mid+1,r]        update(max(s,mid+1),e,2*k+1,x);    }}int ans[N];void caulate_ans(int k){    if(rmq[k].n>0){        for(int i=rmq[k].l;i<=rmq[k].r;++i){            ans[i]+=rmq[k].n;        }    }    if(rmq[k].l<rmq[k].r){        caulate_ans(2*k);        caulate_ans(2*k+1);    }}int main(){    //freopen("/home/lu/文档/r.txt","r",stdin);    //freopen("/home/lu/文档/w.txt","w",stdout);    int n,s,e;    while(scanf("%d",&n),n){        init_rmq(1,n,1);        for(int i=0;i<n;++i){            scanf("%d%d",&s,&e);            update(s,e,1,1);        }        fill(ans,ans+n+1,0);        caulate_ans(1);        printf("%d",ans[1]);        for(int i=2;i<=n;++i){            printf(" %d",ans[i]);        }        putchar('\n');    }    return 0;}