标签：ref else next sum bre jimmy bug ges form
InputThere are several test cases. The first line of each test case contains an integer N (1 <= N <= 200). Each of the next N lines contains a string Si. You can assume the strings contain alphabets (‘a‘-‘z‘, ‘A‘-‘Z‘) only, and the length of every string is no more than 1000.
OutputOutput one line for each test case, indicating the corresponding answer.Sample Input
3
ab
bcc
ccb
1
abcd
Sample Output
6 0
题意：给你n个字符串，n个字符串中两个字符串相连的权值是串1逆置后和串2的最长公共前缀，自身和自身相连的权值为0，问输出最大完美匹配值。
思路：将串1（1~n）从后往前和串2（1~n）进行匹配，匹配值存入两串相连的边（实现：用两个for进行循环匹配），之后就是km算法啦~~
我不知道自己哪里来的想法在读入结束条件加一个n!=0，结果一直超时，我各种优化还是超时，到了饭点才发现自己输入的问题，真是一个浪费时间的bug
#include<stdio.h>
#include<string.h>
#define N 210
#define M 1110
#define INF 0x3f3f3f3f
int w[N][N],lx[N],ly[N];
int visx[N],visy[N],linker[N],slack[N];
int n,ans,nx,ny;
char e[N][M];
void Getmap()//建图
{
int i,j,x,y,count,l;
for(x = 1; x <= n; x ++)
{
for(y = 1; y <= n; y ++)
{
if(x == y)
w[x][y] = 0;
else
{
l = strlen(e[x]+1);
count = 0;
for(i = l,j = 1;i > 0&&e[y][j]!=‘\0‘;i--,j++)
{
if(e[x][i] == e[y][j])
count++;
else
break;
}
w[x][y] = count;
}
}
}
return;
}
int dfs(int x)//寻找增广路
{
int y,tmp;
visx[x] = 1;
for(y = 1; y <= ny; y ++)
{
if(!visy[y])
{
tmp = lx[x] + ly[y] - w[x][y];
if(!tmp)
{
visy[y] = 1;
if(linker[y] == -1||dfs(linker[y]))
{
linker[y] = x;
return 1;
}
}
else if(slack[y] > tmp)
slack[y] = tmp;
}
}
return 0;
}
int KM()//km算法
{
int x,y,i,j,sum,d;
memset(linker,-1,sizeof(linker));
memset(ly,0,sizeof(ly));
for(x = 1; x <= nx; x ++)
for(y = 1,lx[x] = -INF; y <= ny; y ++)
if(lx[x] < w[x][y])
lx[x] = w[x][y];
for(x = 1; x <= nx; x ++)
{
for(i = 1; i <= ny; i ++)
slack[i] = INF;
while(1)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
d = INF;
if(dfs(x))
break;
for(y = 1; y <= ny; y ++)
if(!visy[y]&&d > slack[y])
d = slack[y];
for(i = 1; i <= nx; i ++)
if(visx[i])
lx[i] -= d;
for(i = 1; i <= ny; i ++)
if(visy[i])
ly[i] += d;
else
slack[i]-=d;
}
}
sum = 0;
for(i = 1; i <= ny; i ++)
if(linker[i]!=-1)
sum += w[linker[i]][i];
return sum;
}
int main()
{
int i;
while(scanf("%d",&n)!=EOF)
{
getchar();
nx = ny = n;
for(i = 1; i <= n; i ++)
scanf("%s",e[i]+1);
Getmap();//建图
ans = KM();//km算法
printf("%d\n",ans);
}
return 0;
}
【二分图匹配入门专题1】L - Card Game hdu 3722【km算法】
标签：ref else next sum bre jimmy bug ges form
原文地址：http://www.cnblogs.com/chengdongni/p/7357160.html
