关于KM板子的一个小细节
一个一般不怎么会遇到的小细节。
也是在被下面这道题坑了才发现的。
My Brute
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1932 Accepted Submission(s): 714
Problem Description
Seaco is a beautiful girl and likes play a game called “My Brute”. Before Valentine’s Day, starvae and xingxing ask seaco if she wants to spend the Valentine’s Day with them, but seaco only can spend it with one of them. It’s hard to choose from the two excellent boys. So there will be a competition between starvae and xingxing. The competition is like the game “My Brute”.
Now starvae have n brutes named from S1 to Sn and xingxing’s brutes are named from X1 to Xn. A competition consists of n games. At the beginning, starvae’s brute Si must versus xingxing’s brute Xi. But it’s hard for starvae to win the competition, so starvae can change his brutes’ order to win more games. For the starvae’s brute Si, if it wins the game, starvae can get Vi scores, but if it loses the game, starvae will lose Vi scores. Before the competition, starvae’s score is 0. Each brute can only play one game. After n games, if starvae’s score is larger than 0, we say starvae win the competition, otherwise starvae lose it.
It’s your time to help starvae change the brutes’ order to make starvae’s final score be the largest. If there are multiple orders, you should choose the one whose order changes the least from the original one. The original order is S1, S2, S3 … Sn-1, Sn, while the final order is up to you.
For starvae’s brute Si (maybe this brute is not the original brute Si, it is the ith brute after you ordered them) and xingxing’s brute Xi, at first Si has Hi HP and Xi has Pi HP, Si’s damage is Ai and Xi’s is Bi, in other words, if Si attacks, Xi will lose Ai HP and if Xi attacks, Si will lose Bi HP, Si attacks first, then it’s Xi’s turn, then Si… until one of them’s HP is less than 0 or equal to 0, that, it lose the game, and the other win the game.
Come on, starvae’s happiness is in your hand!
Input
First line is a number n. (1<=n<=90) Then follows a line with n numbers mean V1 to Vn. (0<Vi<1000) Then follows a line with n numbers mean H1 to Hn. (1<=Hi<=100)Then follows a line with n numbers mean P1 to Pn. (1<=Pi<=100) Then follows a line with n numbers mean A1 to An.(1<=Ai<=50) Then follows a line with n numbers mean B1 to Bn. (1<=Bi<=50) A zero signals the end of input and this test case is not to be processed.
Output
For each test case, if starvae can win the competition, print the largest score starvae can get, and then follow a real percentage means the similarity between the original order and the final order you had changed, round it to three digits after the decimal point. If starvae can’t win the competition after changing the order, please just print “Oh, I lose my dear seaco!” Maybe the sample can help you get it.
Sample Input
3
4 5 6
6 8 10
12 14 16
7 7 6
7 3 5
3
4 5 6
6 8 10
12 14 16
5 5 5
5 5 5
0
Sample Output
7 33.333%
Oh, I lose my dear seaco!
Author
starvae
Source
HDU “Valentines Day” Open Programming Contest 2010-02-14
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网上有好几种KM算法的板子。每种板子都有一点点不同。
先来一份AC代码
#include<bits/stdc++.h>
using namespace std;
#define mem(a) memset(a,0,sizeof(a))
const int inf = 0x3f3f3f3f;
int n,a[100],h[100],p[100],v[100],b[100];
int g[100][100];
int cx[100],cy[100],ex[100],ey[100],d;
int vx[100],vy[100];
double rate;
bool dfs(int u)
{
vx[u] = 1;
for (int i = 1;i<=n;i++)
{
if (vy[i]) continue;
int num = ex[u] + ey[i] - g[u][i];
if (num == 0)
{
vy[i] = 1;
if (cy[i] == 0 || dfs(cy[i]))
{
cx[u] = i;
cy[i] = u;
return 1;
}
}
}
return 0;
}
int KM()
{
mem(cx); mem(cy); mem(ey);
for (int i = 1;i<=n;i++)
{
ex[i] = -2147483647;
for (int j = 1;j<=n;j++)
ex[i] = max(ex[i],g[i][j]);
}
for (int i = 1;i<=n;i++)
{
while (1)
{
mem(vx); mem(vy);
if (dfs(i)) break;
d = 2147483647;
for(int j = 1;j <= n;j++)
{
if (vx[j])
{
for (int k = 1;k<=n;k++)
{
if (!vy[k]) d = min(d,ex[j] + ey[k] - g[j][k]);
}
}
}
for(int j = 1;j <= n;j++)
{
if(vx[j]) ex[j] -= d;
if(vy[j]) ey[j] += d;
}
}
}
int res = 0;
rate = 0;
for (int i = 1;i<=n;i++)
{
res += g[cy[i]][i];
if (cy[i] == i) rate++;
}
return res;
}
int main()
{
while(scanf("%d",&n) && (n != 0))
{
for (int i = 1;i<=n;i++) scanf("%d",&v[i]);
for (int i = 1;i<=n;i++) scanf("%d",&h[i]);
for (int i = 1;i<=n;i++) scanf("%d",&p[i]);
for (int i = 1;i<=n;i++) scanf("%d",&a[i]);
for (int i = 1;i<=n;i++) scanf("%d",&b[i]);
for (int i = 1;i<=n;i++)
{
for (int j = 1;j<=n;j++)
{
if ((h[i] / b[j] + (h[i] % b[j] > 0)) >= (p[j] / a[i] + (p[j] % a[i] > 0))) g[i][j] = v[i] * 100;
else g[i][j] = -v[i] * 100;
}
g[i][i]++;
}
int ans = KM() / 100;
if (ans <= 0) printf("Oh, I lose my dear seaco!\n");
else
{
rate = (rate / (double)n) * 100;
printf("%d %.3f%%\n",ans,rate);
}
}
}
有多种版本的找最小的 两边顶标和-边权 的值的方法
我一开始打的板子是这样的
bool dfs(int u)
{
vx[u] = 1;
for (int i = 1;i<=n;i++)
{
if (vy[i]) continue;
int num = ex[u] + ey[i] - g[u][i];
if (num == 0)
{
vy[i] = 1;
if (cy[i] == 0 || dfs(cy[i]))
{
cx[u] = i;
cy[i] = u;
return 1;
}
}
else d = min(d,num);//改变在这里
}
return 0;
}
int KM()
{
mem(cx); mem(cy); mem(ey);
for (int i = 1;i<=n;i++)
{
ex[i] = -2147483647;
for (int j = 1;j<=n;j++)
ex[i] = max(ex[i],g[i][j]);
}
for (int i = 1;i<=n;i++)
{
while (1)
{
mem(vx); mem(vy);
d = 2147483647;
if (dfs(i)) break;
for(int j = 1;j <= n;j++)//还有这里
{
if(vx[j]) ex[j] -= d;
if(vy[j]) ey[j] += d;
}
}
}
int res = 0;
rate = 0;
for (int i = 1;i<=n;i++)
{
res += g[cy[i]][i];
if (cy[i] == i) rate++;
}
return res;
}
上面这个板子在跑dfs的时候,用d记录下了最小的 两边顶标和-边权 的值。
看起来好像没什么问题,
然而我第一份代码46ms跑完了,第二份代码TLE1000ms,调了我半个下午 + 一个晚上终于调出来了。
而且仔细一想第二份代码也没有错。
两者的区别就在于,一个是在
d
f
s
dfs
dfs的时候把
d
d
d找了出来,一个是跑完
d
f
s
dfs
dfs之后,从已经访问了的左边的点和未访问的右边的点的组合中,找出最小的
d
d
d。
在dfs里找会出现一种情况就是 当前
v
y
[
i
]
=
0
vy[i] = 0
vy[i]=0 &&
n
u
m
>
0
num > 0
num>0,然后d把这个值保存下来了,假设它的值是1。然后往下
d
f
s
dfs
dfs的时候,又跑到
i
i
i这个点,这个时候
v
y
[
i
]
=
=
1
vy[i] == 1
vy[i]==1 &&
n
u
m
=
=
0
num == 0
num==0。虽然 这个时候
n
u
m
=
=
0
num == 0
num==0这个值并不会保存到
d
d
d当中,但是
d
d
d里面保存的原来的值1还在。如果是在
d
f
s
dfs
dfs完之后再寻找
d
d
d的话,因为
v
y
[
i
]
=
=
1
vy[i] == 1
vy[i]==1,
d
d
d的值可能就不会是1了。也就是说在
d
f
s
dfs
dfs里找最小值,可能会出现找到的
d
d
d是已经访问了的左边的点和已经访问的右边的点的顶标和减去边权这种情况。所以时间复杂度会上升,所以一直TLE。
本文来自博客园,作者:Un-Defined,转载请保留本文署名Un-Defined,并在文章顶部注明原文链接:https://www.cnblogs.com/EIPsilly/p/17463696.html