hihocoder-1391&&北京网赛09 Countries(优先队列)

题目链接：

Countries

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

There are two antagonistic countries, country A and country B. They are in a war, and keep launching missiles towards each other.

It is known that country A will launch N missiles. The i-th missile will be launched at time Tai. It flies uniformly and take time Taci from one country to the other. Its damage capability is Dai.

It is known that country B will launch M missiles. The i-th missile will be launched at time Tbi.

It flies uniformly and takes time Tbci from one country to the other. Its damage capability is Dbi.

Both of the countries can activate their own defending system.

The defending system of country A can last for time TA, while The defending system of country B can last for time TB.

When the defending system is activated, all missiles reaching the country will turn around and fly back at the same speed as they come.

At other time, the missiles reaching the country will do damages to the country.
(Note that the defending system is still considered active at the exact moment it fails)

Country B will activate its defending system at time X.

When is the best time for country A to activate its defending system? Please calculate the minimal damage country A will suffer.

输入

There are no more than 50 test cases.

For each test case:

The first line contains two integers TA and TB, indicating the lasting time of the defending system of two countries.

The second line contains one integer X, indicating the time that country B will active its defending system.

The third line contains two integers N and M, indicating the number of missiles country A and country B will launch.

Then N lines follow. Each line contains three integers Tai, Taci and Dai, indicating the launching time, flying time and damage capability of the i-th missiles country A launches.

Then M lines follow. Each line contains three integers Tbi, Tbci and Dbi, indicating the launching time, flying time and damage capability of the i-th missiles country B launches.

0 <= TA, TB, X, Tai, Tbi<= 100000000

1 <= Taci, Tbci <= 100000000

0 <= N, M <= 10000

1 <= Dai, Dbi <= 10000

输出

For each test case, output the minimal damage country A will suffer.

提示

In the first case, country A should active its defending system at time 3.

Time 1: the missile is launched by country A.

Time 2: the missile reaches country B, and country B actives its defending system, then the missile turns around.

Time 3: the missile reaches country A, and country A actives its defending system, then the missile turn around.

Time 4: the missile reaches country B and turns around.

Time 5: the missile reaches country A and turns around.

Time 6: the missile reaches country B, causes damages to country B.

样例输入

2 2
2
1 0
1 1 10
4 5
3
2 2
1 2 10
1 5 7
1 3 2
0 4 8

样例输出

0
17

题意：

A,B两国互发导弹,A发n个给了发射时间,到达另一国的时间,和伤害力,B发m个也是一样,
还有就是A,B两国的防御系统的开启时长,B国给了开启时间,现在问A国什么时候打开防御系统才能使伤害最小;

思路:

我们把有可能打到A国的导弹如果要它打在b国需要A国开启的时间段[l,r],然后就转变成了给了一些线段,线段有权值,然后完全覆盖才能取这个权值.然后求最大值;
右端点排序后,枚举左端点,优先队列维护就好了,也算是一个经典模型了;

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e4+10;
int Ta,Tb,x,n,m,st[2*maxn],co[2*maxn],da[2*maxn],g[2*maxn];
struct node
{
    int l,r,da;
}po[2*maxn];
int cmp(node a,node b)
{
    return a.r<b.r;
}
struct MO
{
    int l,da;
    friend bool operator< (const MO &a ,const MO &b)
    {
        return a.l>b.l;
    }
}ha;
priority_queue<MO>qu;
int main()
{
    //freopen("int.txt","r",stdin);
    while(scanf("%d%d",&Ta,&Tb)!=EOF)
    {
        while(!qu.empty())qu.pop();
        scanf("%d",&x);
        scanf("%d%d",&n,&m);
        LL sum=0;
        int cnt=0,u,v,w,num=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            if(u+v>=x&&u+v<=x+Tb)
            {
                st[cnt]=u+v;
                co[cnt]=v;
                da[cnt]=w;
                cnt++;
            }
        }
        for(int i=1;i<=m;i++,cnt++)scanf("%d%d%d",&st[cnt],&co[cnt],&da[cnt]);
        for(int i=0;i<cnt;i++)
        {
            sum=sum+da[i];
            int temp=st[i]+2*co[i];int l=st[i]+co[i],r;
            if(temp>=x&&temp<=Tb+x)
            {
                int len=(x+Tb-temp)/(2*co[i]);
                r=temp+co[i]+len*2*co[i];
                if(r-l>Ta)continue;
                po[++num].l=l;po[num].r=r;po[num].da=da[i];
                g[num]=l;
            }
            else
            {
                r=l;
                po[++num].l=l;po[num].r=r;po[num].da=da[i];
                g[num]=l;
            }
        }
        sort(po+1,po+num+1,cmp);
        sort(g+1,g+num+1);
        LL ans=sum,d=0;
        int fr=1;
        g[0]=-1;
        for(int i=1;i<=num;i++)
        {
            if(g[i]==g[i-1])continue;
            int hi=g[i]+Ta;
            while(po[fr].r<=hi&&fr<=num)
            {
                ha.l=po[fr].l;
                ha.da=po[fr].da;
                d=d+ha.da;
                qu.push(ha);
                fr++;
            }
            while(!qu.empty())
            {
                ha=qu.top();
                if(ha.l<g[i])
                {
                    qu.pop();
                    d=d-ha.da;
                }
                else break;
            }
            ans=min(ans,sum-d);
        }
        printf("%lld\n",ans);
    }
}