346题意: 题目给出n个课程和m天,问如何安排m天的课程使得最后的收益达到最大值.题解: 简单DP,以dp[i]表示当前安排i天可以获得的最大收益,则可以得出状态转移方程dp[k + j] = max(dp[k + j], dp[k] + a[i][j]); 表示当前的课程安排j天的最优解. 具体看下代码吧!AC代码:#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-9using namespace std;typedef long long ll;typedef pair<int,int>P;const int M = 200;const int INF = 0x3f3f3f3f;const int mod = 10000007;int maxn[M][M],dp[M],tp[M];int main(){//freopen("in","r",stdin);int n,m;while(~scanf("%d %d",&n,&m) && (n | m)){memset(dp,0,sizeof(dp));for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){scanf("%d",&maxn[i][j]);}}for(int i = 1; i <= m; i++) dp[i] = maxn[1][i]; //初始化for(int i = 2; i <= n; i++){for(int k = 1; k <= m; k++) tp[k] = dp[k]; //防止当前的状态会覆盖之前的值,所以先把最优值保存在tp数组中;for(int j = 1; j <= m; j++){for(int k = 0; k + j <= m; k++){tp[k + j] = max(tp[k + j],dp[k] + maxn[i][j]);}}for(int k = 1; k <= m; k++) dp[k] = max(tp[k],dp[k]);}int Max = 1;for(int i = 2; i <= m; i++)if(dp[Max] < dp[i]) Max = i;printf("%d\n",dp[Max]);}return 0;}
，不要气馁于那前方的阴影，那只是因为我背后光芒万丈
