hdu 5074 Hatsune Miku(dp)

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题目连接：hdu 5074 Hatsune Miku

题目大意：给定给一个M∗M的矩阵，以及一个长度为N的数组，每次取G[a[i]][a[i+1]]的值，如果a[i]为-1则可以是任何数，问说最大值。

解题思路：dp[i][j]表示第i个数为j的时候的最优解，如果a[i]不为-1，那么j = a[i]，否则枚举j。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105;const int INF = 0x3f3f3f3f;int N, M, g[maxn][maxn], p[maxn], dp[maxn][maxn];void init () {    scanf("%d%d", &N, &M);    memset(g, 0, sizeof(g));    for (int i = 1; i <= M; i++) {        for (int j = 1; j <= M; j++)            scanf("%d", &g[i][j]);    }    for (int i = 1; i <= N; i++)        scanf("%d", &p[i]);}int solve () {    memset(dp, -INF, sizeof(dp));    if (p[1] == -1)        memset(dp[1], 0, sizeof(dp[1]));    else        dp[1][p[1]] = 0;    for (int i = 2; i <= N; i++) {        if (p[i] == -1) {            for (int x = 1; x <= M; x++)                for (int y = 1; y <= M; y++)                    dp[i][x] = max(dp[i][x], dp[i-1][y] + g[y][x]);        } else {            int x = p[i];            for (int y = 1; y <= M; y++)                dp[i][x] = max(dp[i][x], dp[i-1][y] + g[y][x]);        }    }    int ret = 0;    for (int i = 1; i <= M; i++)        ret = max(dp[N][i], ret);    return ret;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        printf("%d\n", solve());    }}

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