您的位置:首页 > 产品设计 > UI/UE

poj2127 Greatest Common Increasing Subsequence(LICS+路径)

2016-03-25 20:09 387 查看
5555校赛在即可是这路径输出我怎么也搞不好,搞了好久了,改天在搞吧。。。


#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <string.h>

using namespace std;

const int N = 1005;
const int INF = 1000005;

int dp

, path

;

int main()
{
freopen("in.txt", "r", stdin);
int a1
, a2
, f
;
int len1, len2;
while(~scanf("%d", &len1))
{
memset(path, 0, sizeof(path));
for(int i = 0; i <= N; i ++)
dp[i][0] = dp[0][i] = 0;
for(int i = 1; i <= len1; i ++)
scanf("%d", &a1[i]);
scanf("%d", &len2);
for(int i = 1; i <= len2; i ++)
scanf("%d", &a2[i]);
int k = 0, endi, endj, ans = 0;
for(int i = 1; i <= len1; i ++)
{
int maxx = 0;
for(int  j = 1; j <= len2; j ++)
{
dp[i][j] = dp[i - 1][j];
if(a1[i] > a2[j])
{
maxx = max(maxx, dp[i][j]);
k = j;
}
if(a1[i] == a2[j])
{
//         printf("<%d>", maxx);
dp[i][j] = maxx + 1;
path[i][j] = k;
printf("<%d, %d, %d>\n", i, j, path[i][j]);
}
if(ans < dp[i][j])
{
ans = max(ans, dp[i][j]);
endi = i;
endj = j;
}
}
}
int sum = ans;
while(sum && endi && endj)
{
if(path[endi][endj] > 0)
{
printf("(%d, %d)", endi, endj);
f[sum --] = path[endi][endj];
//   printf("<%d>", path[endi][endj]);
endj = path[endi][endj] - 1;
}
endi --;
}
//   printf("%d\n", ans);
/*  for(int i = 1; i <= ans; i ++)
printf("%d ", f[i]);*/
for(int i = 1; i < ans; i ++)
printf("%d ", a2[f[i]]);
printf("%d\n", a2[f[ans]]);
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  poj