Devour Magic 線段樹

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Devour MagicTime Limit: 2000ms   Memory limit: 65536K  有疑問？點這裡^_^題目描述In Warcraft III, Destroyer is a large flying unit that must consume magic to sustain its mana. Breaking free of the obsidian stone that holds them, these monstrous creatures roar into battle, swallowing magic to feed their insatiable hunger as they move between battles and rain destruction down upon their foes. Has Spell Immunity. Attacks land and air units.
The core skill of the Destroyer is so called Devour Magic, it takes all mana from all units in a area and gives it to the Destroyer. Now to simplify the problem, assume you have n units in a line, all unit start with 0 mana and can increase to infinity maximum mana. All unit except the Destroyer have mana regeneration 1 in per unit time. The Destroyer have m instructions t l r, it means, in time t, the Destroyer use Devour Magic on unit from l to r. We give you all m instructions in time order, count how many mana the Destroyer have devour altogether.輸入The first line contains one integer T, indicating the test case. For each test case, the first contains two integer n,?m(1?≤?n,?m?≤?10^5). The the next m line each line contains a instruction t l r.(1?≤?t?≤?10^5,?1?≤?l?≤?r?≤?n)輸出For each test case, output the conrespornding result.樣本輸入

110 51 1 102 3 103 5 104 7 105 9 10

樣本輸出

30

提示 來源2014年山東省第五屆ACM大學生程式設計競賽

在1到n這n個單位，每秒都會恢複一滴血，在第t分鐘時，從l到r的這幾個單位血值變為0，求最後變為0的有多少。基本線段樹。

#include<stdio.h>#include<string.h>#include<algorithm>#define M 100007#define inf 0x3f3f3f3fusing namespace std;long long sum;long long r,c,m;struct T{    long long left,right,sum;    long long add,set;}tree[M<<4];void build(long long l,long long r,long long i){    tree[i].left=l;tree[i].right=r;tree[i].set=-1;    tree[i].add=tree[i].sum=0;    if(l==r)return;    long long mid=l+(r-l)/2;    build(l,mid,i<<1);    build(mid+1,r,i<<1|1);}void pushdown(long long i){    if(tree[i].left>=tree[i].right)return;    if(tree[i].set!=-1)    {        tree[i<<1].set=tree[i<<1|1].set=tree[i].set;        tree[i<<1].add=tree[i<<1|1].add=0;        tree[i<<1].sum=(tree[i<<1].right-tree[i<<1].left+1)*tree[i].set;        tree[i<<1|1].sum=(tree[i<<1|1].right-tree[i<<1|1].left+1)*tree[i].set;    }    if(tree[i].add>0)    {        long long add=tree[i].add;        tree[i<<1].add+=add;tree[i<<1|1].add+=add;        tree[i<<1].sum+=add*(tree[i<<1].right-tree[i<<1].left+1);        tree[i<<1|1].sum+=add*(tree[i<<1|1].right-tree[i<<1|1].left+1);    }}void maintain(long long i){    if(tree[i].left>=tree[i].right)return;    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;}void update_add(long long l,long long r,long long val,long long i){    if(tree[i].left==l&&tree[i].right==r)    {        tree[i].add+=val;        tree[i].sum+=(tree[i].right-tree[i].left+1)*val;        return;    }    pushdown(i);    tree[i].set=-1;    tree[i].add=0;    long long mid=tree[i].left+(tree[i].right-tree[i].left)/2;    if(r<=mid)update_add(l,r,val,i<<1);    else if(l>mid)update_add(l,r,val,i<<1|1);    else    {        update_add(l,mid,val,i<<1);        update_add(mid+1,r,val,i<<1|1);    }    maintain(i);}void update_set(long long l,long long r,long long val,long long i){    if(tree[i].left==l&&tree[i].right==r)    {        tree[i].set=val;        tree[i].add=0;        tree[i].sum=(tree[i].right-tree[i].left+1)*val;        return;    }    pushdown(i);    tree[i].set=-1;    tree[i].add=0;    long long mid=tree[i].left+(tree[i].right-tree[i].left)/2;    if(r<=mid)update_set(l,r,val,i<<1);    else if(l>mid)update_set(l,r,val,i<<1|1);    else    {        update_set(l,mid,val,i<<1);        update_set(mid+1,r,val,i<<1|1);    }    maintain(i);}void query(long long l,long long r,long long i){    if(tree[i].left==l&&tree[i].right==r)    {       sum+=tree[i].sum;        return;    }    pushdown(i);    tree[i].set=-1;tree[i].add=0;    long long mid=tree[i].left+(tree[i].right-tree[i].left)/2;    if(r<=mid)query(l,r,i<<1);    else if(l>mid)query(l,r,i<<1|1);    else    {        query(l,mid,i<<1);        query(mid+1,r,i<<1|1);    }    maintain(i);}int main(){    long long t;    scanf("%lld",&t);    while(t--)    {        long long n,m;        scanf("%lld%lld",&n,&m);        build(1,n,1);        long long a,b,c,time=0,summ=0;        while(m--)        {            sum=0;            scanf("%lld%lld%lld",&c,&a,&b);            update_add(1,n,c-time,1);            time=c;            query(a,b,1);            summ+=sum;            update_set(a,b,0,1);        }        printf("%lld\n",summ);    }    return 0;}

Devour Magic 線段樹