Hatsune MikuTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 654Accepted Submission(s): 471Problem Description
Hatsune Miku is a popular virtual singer. It is very popular in both Japan and China. Basically it is a computer software that allows you to compose a song on your own using the vocal package.Today you want to compose a song, which is just a sequence of notes. There are only m different notes provided in the package. And you want to make a song with n notes.
Also, you know that there is a system to evaluate the beautifulness of a song. For each two consecutive notes a and b, if b comes after a, then the beautifulness for these two notes is evaluated as score(a, b).So the total beautifulness for a song consisting of notes a1, a2, . . . , an, is simply the sum of score(ai, ai+1) for 1 ≤ i ≤ n – 1.Now, you find that at some positions, the notes have to be some specific ones, but at other positions you can decide what notes to use. You want to maximize your song’s beautifulness. What is the maximum beautifulness you can achieve?
Input
The first line contains an integer T (T ≤ 10), denoting the number of the test cases.For each test case, the first line contains two integers n(1 ≤ n ≤ 100) and m(1 ≤ m ≤ 50) as mentioned above. Then m lines follow, each of them consisting of m space-separated integers, the j-th integer in the i-th line for score(i, j)( 0 ≤ score(i, j) ≤ 100). The next line contains n integers, a1, a2, . . . , an (-1 ≤ ai ≤ m, ai ≠ 0), where positive integers stand for the notes you cannot change, while negative integers are what you can replace with arbitrary notes. The notes are named from 1 to m.
Output
For each test case, output the answer in one line.
Sample Input
25 383 86 7715 93 3586 92 493 3 3 1 210 536 11 68 67 2982 30 62 23 6735 29 2 22 5869 67 93 56 1142 29 73 21 19-1 -1 5 -1 4 -1 -1 -1 4 -1
Sample Output
270625
Source
题目链接：?pid=5074题目大意：输入一个m*m的矩阵map[][]和一个长度为n的序列a1a2…an，则结果为map[a1][a2]+map[a2][a3]+…+map[an-1][an]，问题是序列里可能有－1，－1可以代表1-m中的任何数，现在问怎样选择序列，可以使最后的结果值最大，输出最大值题目分析：线性dp，dp[i][j]表示序列的第i个位置上放的是数字j，则分成4种情况1）a[i] > 0 && a[i – 1] > 0，，此时两个点位置都确定直接加即可2）a[i] > 0 && a[i – 1] <= 0，此时前一点位置不确定，枚举前一点位置3）a[i] <= 0 && a[i – 1] > 0，此时后一点位置不确定，枚举后一点位置4）a[i] <= 0 && a[i – 1] <= 0，前后都不确定，分别枚举前后两点
最后答案在dp[n][i]中取大，(1 <= i <= m)表示第n个位置放1-m 中的最大值#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int dp[105][55], map[55][55], a[105];int main(){int T;scanf("%d", &T);while(T–){int n, m;scanf("%d %d", &n, &m);for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)scanf("%d ", &map[i][j]);for(int i = 1; i <= n; i++)scanf("%d", &a[i]);memset(dp, 0, sizeof(dp));for(int i = 2; i <= n; i++){if(a[i] > 0){if(a[i – 1] > 0)dp[i][a[i]] = dp[i – 1][a[i – 1]] + map[a[i – 1]][a[i]];elsefor(int j = 1; j <= m; j++)dp[i][a[i]] = max(dp[i][a[i]], dp[i – 1][j] + map[j][a[i]]);}else{for(int j = 1; j <= m; j++){if(a[i – 1] > 0)dp[i][j] = dp[i – 1][a[i – 1]] + map[a[i – 1]][j];elsefor(int k = 1; k <= m; k++)dp[i][j] = max(dp[i][j], dp[i – 1][k] + map[k][j]);}}}int ans = 0;for(int i = 1; i <= m; i++)ans = max(ans, dp[n][i]);printf("%d\n", ans);}}
而它的种子，就是它生命的延续，继续承受风，
