Ministry(Ural) DP Hqg_ac

这道题目其实就是一个裸的$Dijkstra$，但我在这里讲一下DP怎么求解

$dp[i][j]$表示第$i$行第$j$列的最小值
$dp[i][j]=min(dp[i-1][j],dp[i][j±1])$ 这个转移出现环，不可行
改进:
我们发现对于$dp[i][j]$，在一次循环中，他只有可能被一个方向的dp更新（另一个方向只会比他大）
例如:当j从2枚举到$n$时 $dp[i][j]$只会被$dp[i-1][j]$和$dp[i][j-1]$更新
当j从$n-1$枚举到1时 $dp[i][j]$只会被$dp[i-1][j]$和$dp[i][j+1]$更新
提炼出$dp[i-1][j]$，即可得到转移方法：
$dp[i][j]=${ $dp[i-1][j];$
$for (j=2->n) min(dp[i][j-1])$
$for (j=n-1->1) min(dp[i][j-1])$
}
这个转移不会出现环，可行，时间复杂度$O({{n}^{2}})$

code:

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int M, N, cost[105][505];
long long dp[105][505];
int path[105*505];

int DP() { 
    for (int i = 1; i <= N; ++i) {
        dp[1][i] = cost[1][i]; 
        path[N+i] = 0;
    }
    for (int i = 2; i <= M; ++i) {
        for (int j = 1; j <= N; ++j) { 
            dp[i][j] = dp[i-1][j] + cost[i][j];
            path[i*N+j] = (i-1)*N+j;

        }
        for (int j = 2; j <= N; ++j) { 
            if (dp[i][j] > dp[i][j-1] + cost[i][j]) {
                dp[i][j] = dp[i][j-1] + cost[i][j];
                path[i*N+j] = i*N+j-1;
            }
        }
        for (int j = N-1; j >= 1; --j) {  
            if (dp[i][j] > dp[i][j+1] + cost[i][j]) {
                dp[i][j] = dp[i][j+1] + cost[i][j];
                path[i*N+j] = i*N+j+1;
            }
        }
    }
    int Min = dp[M][1], end = 1;
    for (int i = 2; i <= N; ++i) {
        if (Min > dp[M][i]) {
            Min = dp[M][i];
            end = i;
        }
    }
    return M*N+end;
}

void output(int x) {
    if (path[x]) {
        output(path[x]);
    }
    if (x % N != 0) {
        printf("%d ", x-N*(x/N)); //如果是提交POJ或Open_Bailian的请把' '->\n
    }
    else {
        printf("%d ", N); //如果是提交POJ或Open_Bailian的请把' '->\n
    }
}

int main() {
    int pos;
    while (scanf("%d %d", &M, &N) == 2) {
        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                scanf("%d", &cost[i][j]);
            }
        }
        pos = DP(); 
        output(pos);
    }
    return 0;    
}