UVALive 7146Defeat the Enemy贪心

Long long ago there is a strong tribe living on the earth. They always have wars and eonquer others.One day, there is another tribe become their target. The strong tribe has decide to terminate them!!!There are m villages in the other tribe. Each village contains a troop with attack power EAttacki,and defense power EDefensei. Our tribe has n troops to attack the enemy. Each troop also has theattack power Attacki, and defense power Defensei. We can use at most one troop to attack one enemyvillage and a troop can only be used to attack only one enemy village. Even if a troop survives anattack, it can’t be used again in another attack.The battle between 2 troops are really simple. The troops use their attack power to attack againstthe other troop simultaneously. If a troop’s defense power is less than or equal to the other troop’sattack power, it will be destroyed. It’s possible that both troops survive or destroy.The main target of our tribe is to destroy all the enemy troops. Also, our tribe would like to havemost number of troops survive in this war.InputThe first line of the input gives the number of test cases, T. T test cases follow. Each test case startwith 2 numbers n and m, the number of our troops and the number of enemy villages. n lines follow,each with Attacki and Defensei, the attack power and defense power of our troops. The next mlines describe the enemy troops. Each line consist of EAttacki and EDefensei, the attack power anddefense power of enemy troopsOutputFor each test ease, output one line containing ‘Case #x: y’, where x is the test case number (startingfrom 1) and y is the max number of survive troops of our tribe. If it‘s impossible to destroy all enemytroops, output ‘-1’ instead.Limits:1 ≤ T ≤ 100,1 ≤ n, m ≤ 105,1 ≤ Attacki, Defensei, EAttacki, EDefensei ≤ 109,Sample Input23 25 77 31 24 42 22 13 41 105 6Sample OutputCase #1: 3Case #2: -1

题意：我方有n个军队，敌方有m个军队，每一个军队有攻击力a[i].l和防御力a[i].r,我方的每一个军队只能攻击一个敌方的军队，且只能攻击一次，当攻击的军队的攻击力大于等于敌方的防御力，若防御力大于等于敌方的攻击力，那么敌方军队消灭，且我方攻击的军队还存在，否则我两个军队同归于尽，问要把敌方的m个军队都消灭，我方至少损失多少军队。

思路：我方军队按攻击力递减排序，敌方按防御力递减排序。

依次枚举敌人，把我方攻击力大于等于地方防御力的加入multiset，然后找set中有没有防御力大于敌方攻击力的，有的话直接删去我方的这个部队，否则删掉集合中最后一个元素。

代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;using namespace std;typedef struct troop{    int s,t;    bool operator<(const troop & a)const    {        return s>a.s;    }}T;int cmp(const troop &a,const troop &b){    return a.s>b.s;}int cmp2(const troop &a,const troop &b){    return a.t>b.t;}multiset<int>ms;int main(){    int i,j,k,p,t;    int n,m,flag;    T f[100001],r[100001];    cin>>t;    for(p=1;p<=t;p++)    {        cin>>n>>m;        for(i=0;i<n;i++)            scanf("%d%d",&f[i].s,&f[i].t);        for(i=0;i<m;i++)            scanf("%d%d",&r[i].s,&r[i].t);        if(m>n)            cout<<"Case #"<<p<<": -1"<<endl;        else        {            int ans=n;            flag=1;            sort(f,f+n,cmp);            sort(r,r+m,cmp2);            ms.clear();            j=0;            for(i=0;i<m;i++)            {                for(;j<n;j++)                {                    if(f[j].s>=r[i].t)                        ms.insert(f[j].t);                    else break;                }                if(ms.empty())                {                    flag=0;                    break;                }                auto it=ms.upper_bound(r[i].s);                if(it==ms.end())                {                    ans--;                    it=ms.begin();                    ms.erase(it);                }                else ms.erase(it);            }            if(!flag)cout<<"Case #"<<p<<": -1"<<endl;            else cout<<"Case #"<<p<<": "<<ans<<endl;        }    }    return 0;}