命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8488 Accepted Submission(s): 2991 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了!可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知道,…
命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8600 Accepted Submission(s): 3032 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了! 可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知…
Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了!可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知道,不论何人,若在迷宫中被困1小时以上,则必死无疑!可怜的yifenfei为了去救MM,义无返顾地跳进了迷宫.让我们一起帮帮执着的他吧!命运大迷宫可以看成是一个两维的方格阵列,如下图所示: yifenfei一开始在左上角,目的当然是到达右下角的大魔王所在地.迷宫的每一个格子都受到幸运女神眷恋…
命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14555 Accepted Submission(s): 5119 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了!可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知道…
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2571 命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13394 Accepted Submission(s): 4686 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了!可谁能想到,yif…
命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16889 Accepted Submission(s): 5888 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了! 可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要…
命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6674 Accepted Submission(s): 2349 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了!可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知道,…
命运 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15663 Accepted Submission(s): 5486 Problem Description 穿过幽谷意味着离大魔王lemon已经无限接近了! 可谁能想到,yifenfei在斩杀了一些虾兵蟹将后,却再次面临命运大迷宫的考验,这是魔王lemon设下的又一个机关.要知…
DP. /* 2571 */ #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 25 #define MAXM 1005 #define INF -999999 int map[MAXN][MAXM]; int dp[MAXN][MAXM]; int max(int a, int b) { return a>b ? a:b; } int main() { int t, n, m…
简单dp 状态方程很好想,主要是初始化.... 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> using namespace std; #define MAX 1010 #define _INF -100000000 int n,m; int f[MAX][MAX]; int dp[MAX][MAX]; int main() { int t; scanf…