sgu 104 Little shop of flowers

DP

题意很简单， 直接DP就可以了， 打印路径的时候我是从后向前遍历，每次dp[i][j-1]< dp[i][j]时说明发生了最近一次的更新，记录下来然后i--，到上一个花瓶的放置情况。

/*Author :usedrose  */
/*Created Time :2015/7/25 13:48:40*/
/*File Name :2.cpp*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <vector>
#include <string>
#include <ctime>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
#define MAXN 110
#define OK cout << "ok" << endl;
#define o(a) cout << #a << " = " << a << endl
#define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
using namespace std;
typedef long long LL;

int n, m;
int dp[MAXN][MAXN];
int G[MAXN][MAXN];

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    cin.tie(0);
    ios::sync_with_stdio(false);
    while (cin >> n >> m) {
        for (int i = 1;i <= n; ++ i)
            for (int j = 1;j <= m; ++ j)
                cin >> G[i][j];
        for (int i = 0;i <= n; ++ i)
            for (int j = 0;j <= m; ++ j)
                dp[i][j] = -INF;
        for (int i = 0;i <= m; ++ i)
            dp[0][i] = 0;
        for (int i = 1;i <= n; ++ i)
            for (int j = 1;j <= m; ++ j) {
                if (i == 1)
                    dp[i][j] = max(dp[i][j-1], G[i][j]);
                else {
                    dp[i][j] = max(dp[i][j-1], dp[i-1][j-1] + G[i][j]);
                }
            }
        cout << dp[n][m] << endl;
        stack<int> stk;
        int i = n;
        for (int j = m;j >= 1;-- j) {
            if (i == 0) break;
            if (dp[i][j-1] < dp[i][j]) {
                stk.push(j);
                i--;
            }
        }
        cout << stk.top();
        stk.pop();
        while (!stk.empty()) {
            int t = stk.top();
            stk.pop();
            cout << " " << t;
        }
        cout << endl;
    }
    return 0;
}