【SGU】SGU每日练1·Little shop of flowers【DP】

题目大意：

给你n*m的矩形（m >= n）

每个节点mp[i][j]有一个权值，从第一行走到最后一行，每一行只准选择一个数且对于i行，所选数的列数要严格大于i-1行选择的列数

问你最大权值是多少，并输出选择的n个列数

思路：

DP方程非常好想：DP[i][j] = max(DP[i][j - 1], DP[i - 1][j - 1] + mp[i][j]);

找路径的话，可以每行可以从从i+1到m，也可以直接从i - 1开始找

也可以在DP里面做标记，状态转移的时候将此点记录！

但是不能想的太无脑，你懂的

附上代码;

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#pragma warning(disable:4996)

#define Zero(a) memset(a, 0, sizeof(a))
#define Neg(a)  memset(a, -1, sizeof(a))
#define All(a) a.begin(), a.end()
#define PB push_back
#define repf(i,a,b) for(int i = a;i <= b; i++)
#define repff(i,a,b) for(int i = a; i < b; ++i)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,n,1
#define ld rt << 1
#define rd rt << 1 | 1
#define ll long long
#define MAXN 105
#define INF 0x3f3f3f3f
#define mod 10007
using namespace std;
vector<int>go;
int n, m, dp[MAXN][MAXN], gg[MAXN][MAXN], mp[MAXN][MAXN];
void init(){
    memset(gg, 0, sizeof(gg));
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j){
            scanf("%d", &mp[i][j]);
            if(i != 0)dp[i][j] = -INF;
            else dp[i][j] = 0;
        }
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= m; ++j){
            if (i != 0)dp[i][j] = -INF;
            else dp[i][j] = 0;
        }
        go.clear();
}

int main(){
    while (~scanf("%d%d", &n, &m)){
        init();
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j){
                if (dp[i - 1][j - 1] + mp[i][j] > dp[i][j - 1]){
                    gg[i][j] = 1;
                    //cout << "go" << endl;
                    dp[i][j] = dp[i - 1][j - 1] + mp[i][j];
                }
                else{
                    dp[i][j] = dp[i][j - 1];
                }
            }
        printf("%d\n", dp[n][m]);
        int nowx = n,nowy = m;
        while (nowx && nowy){
            if (gg[nowx][nowy] == 1){
                go.push_back(nowy);
                nowx--;
                nowy--;
            }
            else{
                nowy--;
            }
        }
        int len = go.size();
        for (int i = len - 1; i >= 0; --i){
            if (i != len - 1) printf(" ");
            printf("%d", go[i]);
        }
        puts("");
    }
    return 0;
}