STRENGTH
题目链接
题目描述
Strength gives you the confidence within yourself to overcome any fears, challenges or doubts. Feel the fear and do it anyway! If you have been going through a rough time and feel burnt out or stressed, the Strength card encourages you to find the strength within yourself and keep going. You have got what it takes to see this situation through to its eventual end. You might also feel compelled to hold space for someone else who is going through a difficult period and needs your strength and support.
Alice and Bob are playing ``Yu-Gi-Oh!'', a famous turn-based trading card game, in which two players perform their turns alternatively. After several turns, Alice and Bob have many monsters respectively. Alice has n and Bob has m monsters under their own control. Each monster's strength is measured by a non-negative integer si. To be specific, the larger si is, the more power the monster has.
During each turn, for every single monster under control, the player can give a command to it at most once, driving it to battle with an enemy monster (given that opposite player has no monsters as a shield, the monster can directly attack him).
Additionally, the process of the battle is also quite simple. When two monsters battle with each other, the stronger one (i.e. the one with larger si) will overwhelm the other and destroy it and the winner's strength will remain unchanged. Meanwhile, the difference of their strength will produce equivalent damage to the player who loses the battle. If the player is directly attacked by a monster, he will suffer from the damage equal to the monster's strength. Notice that when two monsters have the same strength, both of them will vanish and no damage will be dealt.
Right now it is Alice's turn to play, having known the strength of all monsters, she wants to calculate the maximal damage she can deal towards Bob in one turn. Unfortunately, Bob has great foresight and is well-prepared for the upcoming attack. Bob has converted several of his monsters into defense position, in which even if the monster is destroyed, he wouldn't get any damage.
Now you are informed of the strength of all the monsters and whether it is in defense position for each Bob's monster, you are expected to figure out the maximal damage that could be dealt in this turn.
输入
The first line contains a single integer T≤20 indicating the number of test cases.
For each test case, the first line includes two integer O≤n,m≤100000, representing the number of monsters owned by Alice and Bob.
In next three lines, the first two lines include n and m integers O≤si≤109 indicating the strength of the i-th monster, separated by spaces. The last line contains m integers 0 or 1 indicating the position of Bob’Si-th monsters. In other words, 0 represents the normal position and 1 represents the defense position.
输出
For the ith test, output a single line in beginning of ``Case i:'', followed by an integer indicating the answer, separated by a single space.
样例输入
2
4 2
10 10 10 20
5 15
0 1
4 2
10 10 10 20
5 25
0 1
样例输出
Case 1: 25
Case 2: 15
题意
玩游戏王牌，我方场上n只怪兽，都是攻击形态，战斗力为a[i]；
敌方m只怪兽，战斗力为b[i]，如果下一行是0表示攻击形态，1表示防御形态
如果我方怪兽的战斗力攻击对方攻击形态的怪兽，就会消灭对方的怪兽，并造成超杀（对敌人本体造成战斗力的差值a[i] - b[i]的血量），如果攻击防御形态的怪兽兽，则会消灭怪兽而不会扣对方本体的血量
如果敌人没有怪兽了，就可以对敌人本体造成直接伤害a[i]；
每只怪兽只能攻击一次
问怎么才能扣敌人最多的血
题解
有两种贪心策略
一种是 直接拿战斗力高的去怼敌人战斗力低的攻击形态怪，这样造成的超杀值高
第二种 用战斗力高的去消灭所有的防御怪和攻击怪，然后直接对本体进行攻击
分别求出两种贪心策略的结果求max
代码写的相当繁琐
代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define scac(x) scanf("%c",&x)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define ll long long
#define LL long long
#define read read()
#define pb push_back
#define mp make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int maxn = 1e5+5;
ll a[maxn],b[maxn];
ll g[maxn];//gong
ll s[maxn];//shou
int vis[maxn];
int op;
bool cmp(ll x,ll y)
{
return x > y;
}
int main()
{
int t;
sca(t);
int kase = 0;
while(t--)
{
memset(vis,0,sizeof(vis));
int n,m;
sca2(n,m);
rep(i,0,n) scl(a[i]);
rep(i,0,m) scl(b[i]);
int cntg = 0;
int cnts = 0;
rep(i,0,m)
{
sca(op);
if(op) s[cnts++] = b[i];
else g[cntg++] = b[i];
}
sort(a,a+n,cmp); // da -> xiao
sort(g,g+cntg); // xiao -> da
sort(s,s+cnts,cmp); // da -> xiao
int posa = 0;
int posg = 0;
int poss = cnts-1;
ll ans = -1;
ll temp = 0;
while(posa < n && posg < cntg) //plan1 先打攻击怪
{
if(a[posa] >= g[posg])
{
temp += a[posa] - g[posg];
posa++;
posg++;
}
else
break;
}
ll sum = 0;
if(posg != cntg) //我方怪打完了，对方攻击怪还有剩余
{
ans = max(ans, temp);
}
else
{
int i = n-1;
while(i >= posa && poss >= 0)//开始打防守怪
{
if(a[i] >= s[poss])
{
i--;
poss--;
}
else
{
sum += a[i];
i--;
}
}
if(poss == -1)
{
temp += sum;
if(i != posa)
{
while(i>=posa)
{
temp += a[i];
i--;
}
}
}
ans = max(ans,temp);
}
posa = n-1;
poss = cnts - 1;
while(posa >= 0 && poss >= 0) //plan2 先打防守怪
{
if(a[posa] >= s[poss])
{
vis[posa] = 1;
posa--;
poss--;
}
else
{
posa--;
}
}
if(poss == -1)
{
posa = 0;
posg = cntg - 1;
temp = 0;
while(posa < n && posg >= 0)
{
if(vis[posa]) //前面拿来打防守怪了
{
posa++;
continue;
}
if(a[posa] >= g[posg])
{
temp += a[posa] - g[posg];
posa++;
posg--;
}
else
{
break;
}
}
if(posg == -1)
{
while(posa < n)
{
if(vis[posa])
{
posa++;
continue;
}
temp += a[posa];
posa++;
}
ans = max(ans,temp);
}
}
printf("Case %d: %lld
",++kase,ans);
}
return 0;
}