国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

POJ3280-Cheapest Palindrome-區間dp

2019-11-08 18:25:31
字體:
來源:轉載
供稿:網友

原題鏈接 Cheapest Palindrome Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9310 Accepted: 4494 Description

Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow an electronic ID tag that the system will read as the cows pass by a scanner. Each ID tag’s contents are currently a single string with length M (1 ≤ M ≤ 2,000) characters drawn from an alphabet of N (1 ≤ N ≤ 26) different symbols (namely, the lower-case roman alphabet).

Cows, being the mischievous creatures they are, sometimes try to spoof the system by walking backwards. While a cow whose ID is “abcba” would read the same no matter which direction the she walks, a cow with the ID “abcb” can potentially register as two different IDs (“abcb” and “bcba”).

FJ would like to change the cows’s ID tags so they read the same no matter which direction the cow walks by. For example, “abcb” can be changed by adding “a” at the end to form “abcba” so that the ID is palindromic (reads the same forwards and backwards). Some other ways to change the ID to be palindromic are include adding the three letters “bcb” to the begining to yield the ID “bcbabcb” or removing the letter “a” to yield the ID “bcb”. One can add or remove characters at any location in the string yielding a string longer or shorter than the original string.

Unfortunately as the ID tags are electronic, each character insertion or deletion has a cost (0 ≤ cost ≤ 10,000) which varies depending on exactly which character value to be added or deleted. Given the content of a cow’s ID tag and the cost of inserting or deleting each of the alphabet’s characters, find the minimum cost to change the ID tag so it satisfies FJ’s requirements. An empty ID tag is considered to satisfy the requirements of reading the same forward and backward. Only letters with associated costs can be added to a string.

Input

Line 1: Two space-separated integers: N and M Line 2: This line contains exactly M characters which constitute the initial ID string Lines 3..N+2: Each line contains three space-separated entities: a character of the input alphabet and two integers which are respectively the cost of adding and deleting that character. Output

Line 1: A single line with a single integer that is the minimum cost to change the given name tag. Sample Input

3 4 abcb a 1000 1100 b 350 700 c 200 800 Sample Output

900 Hint

If we insert an “a” on the end to get “abcba”, the cost would be 1000. If we delete the “a” on the beginning to get “bcb”, the cost would be 1100. If we insert “bcb” at the begining of the string, the cost would be 350 + 200 + 350 = 900, which is the minimum. Source

USACO 2007 Open Gold

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int maxn=2000 + 10;const int INF = 0x3f3f3f3f;int dp[maxn][maxn],cost[1000],n,m,cost1,cost2;int main(){ cin >> n >> m; char s[maxn],ch; cin >> s; for(int i=m;i>0;i--) s[i]=s[i-1]; while(n--){ cin >> ch >> cost1 >> cost2; cost[ch] = min(cost1,cost2); } for(int j=1;j<m;j++){ for(int i=1;i<=m-j;i++){ int ni=i,nj=i+j; dp[ni][nj]=INF; if(s[ni]==s[nj]) dp[ni][nj]=dp[ni+1][nj-1]; dp[ni][nj] = min(dp[ni][nj],dp[ni+1][nj] + cost[s[ni]]); dp[ni][nj] = min(dp[ni][nj],dp[ni][nj-1] + cost[s[nj]]); } } cout << dp[1][m] << endl;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 万山特区| 沿河| 六盘水市| 滁州市| 平和县| 开封县| 吴堡县| 留坝县| 沁源县| 虎林市| 保德县| 军事| 含山县| 万州区| 舞钢市| 堆龙德庆县| 义乌市| 绥中县| 原阳县| 临颍县| 崇礼县| 开化县| 泾阳县| 隆尧县| 黄山市| 信丰县| 谷城县| 长沙县| 洛川县| 永城市| 平罗县| 赫章县| 吉安县| 扎赉特旗| 五原县| 民权县| 静乐县| 河北区| 永寿县| 海阳市| 阿鲁科尔沁旗|