hdu 5092 Seam Carving(dp)

题目链接：hdu 5092 Seam Carving

题目大意：给定一个N∗M的矩阵，每次可以向下面三个方向移动，现在要求走过的路径上的值最小，并且尽量靠右。

输出值和路径。

解题思路：dp[i][j]表示移动到第i层j的位置的最优解，r[i][j]则对应记录的是从上一层的哪一个位置到来。水题，只是坑题

意。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 105;
const int inf = 0x3f3f3f3f;
int N, M, g[maxn][maxn], dp[maxn][maxn], vi[maxn][maxn];

void init () {
    scanf("%d%d", &N, &M);
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= M; j++)
            scanf("%d", &g[i][j]);
// memset(vi, -1, sizeof(vi));
    memset(dp, 0, sizeof(dp));
}

void solve () {

    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++) {
            int x = dp[i-1][j], y = j;
            if (j > 1 && dp[i-1][j-1] < x) {
                x = dp[i-1][j-1];
                y = j-1;
            }

            if (j < M && dp[i-1][j+1] <= x) {
                x = dp[i-1][j+1];
                y = j+1;
            }

            dp[i][j] = x + g[i][j];
            vi[i][j] = y;
        }
    }

    int ans = dp[N][1], idx = 1, pos[maxn];
    for (int i = 2; i <= M; i++) if (dp[N][i] <= ans) {
        ans = dp[N][i];
        idx = i;
    }

    for (int i = N; i; i--) {
        pos[i] = idx;
        idx = vi[i][idx];
    }
    for (int i = 1; i <= N; i++)
        printf("%d%c", pos[i], i == N ? '\n' : ' ');
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        printf("Case %d\n", kcas);
        init();
        solve();
    }

    return 0;
}