HDU 5074 Hatsune Miku （dp）

题意:

给你一些音符之间的联系，给你一个串，让你求出这个串的最大值。−1的时候可以任意替代，其他情况必须为序列上的数

分析:

dp[i][j]:=前i个音符,以j类音符结尾的串的最大值,然后转移就好了

代码:

////  Created by TaoSama on 2015-10-10//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, a[55][55], b[105];int dp[105][55];void getMax(int &x, int y) {    x = max(x, y);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d%d", &n, &m);        for(int i = 1; i <= m; ++i)            for(int j = 1; j <= m; ++j)                scanf("%d", &a[i][j]);        for(int i = 1; i <= n; ++i) scanf("%d", b + i);        memset(dp, 0, sizeof dp);        for(int i = 1; i < n; ++i) {            if(b[i] < 0) {                for(int j = 1; j <= m; ++j) {                    if(b[i + 1] < 0) {                        for(int k = 1; k <= m; ++k)                            getMax(dp[i + 1][k], dp[i][j] + a[j][k]);                    } else getMax(dp[i + 1][b[i + 1]], dp[i][j] + a[j][b[i + 1]]);                }            } else {                if(b[i + 1] < 0) {                    for(int k = 1; k <= m; ++k)                        getMax(dp[i + 1][k], dp[i][b[i]] + a[b[i]][k]);                } else getMax(dp[i + 1][b[i + 1]], dp[i][b[i]] + a[b[i]][b[i + 1]]);            }        }        int ans = 0;        for(int i = 1; i <= m; ++i) getMax(ans, dp[n][i]);        printf("%d\n", ans);    }    return 0;}