[轉自qywyh]

PKU 3093 Margaritas on the River Walk
        先對輸入的數組排序，然后類似于01對a[i]做決策，核心代碼加了注釋：
         for (i=1; i<=n; i++) {
                 for (j=1; j<=maxsum; j++) {
                        if (j >= sum[i]) d[i][j] = 1; //j比sum[i]大，肯定這時候d[i][j]=1;
                        else {
                                d[i][j] = d[i-1][j];//不考慮a[i]
                                if (j-a[i]>=0) {//考慮a[i]
                                         if (d[i-1][j-a[i]] > 0) d[i][j] += d[i-1][j-a[i]];//把a[i]加進以前的選擇里面
                                         else d[i][j]++;//a[i]單獨作為一個選擇(這里需要先對a[i]排序，消除后效性)
                               }
                        }
                 }
         }

PKU 1037 A decorative fence
        先dp算出以i為起點的序列的個數，再組合數學
        td[n][i]和tu[n][i]分別表示個數為n，以i開始的上升和下降的序列個數
        易知:
        td[n][1] = 0;
        td[n][i] = sigma(tu[n-1][j], j從1..i-1)  = td[n][i-1] + tu[n-1][i-1] ;
        tu[n][i]  = td[n][n+i-1];

PKU 2677 Tour
        雙調歐幾里德旅行商問題(明顯階段dp)
        動態規劃方程 ：d[i+1][i] = mint(d[i+1][i], d[i][j]+g[j][i+1]); 
                                      d[i+1][j] = mint(d[i+1][j], d[i][j]+g[i][i+1]);
                                       0<=j<i   

PKU 2288 Islands and Bridges
        集合DP
        狀態表示: d[i][j][k] (i為13為二進制表示點的狀態, j為當前節點, k為到達j的前驅節點)