poj-1015(状态转移的方向（01背包）和结果的输出）

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <cstdio>
 using namespace std;
 int dp[][];
 //int path[25][1000];
 vector<int> path[][];
 int sub[];
 int add[];
 int n,m,fix;
 int main ()
 {
     int num=;
     while (cin>>n>>m&&n&&m)  {
         memset (dp,-,sizeof(dp));
         for (int i=;i<=m;i++)
             for (int j=;j<;j++)
                 path[i][j].clear();
         for (int i=;i<=n;i++) {
             int d,p; cin>>d>>p;
             sub[i]=d-p;
             add[i]=d+p;
         }
         fix=*m;  dp[][fix]=;
         for (int i=;i<=n;i++) {
             for (int x=m-;x>=;x--)
                 for (int y=;y<=*fix;y++) {//因为sub[i]既可以为正也可以为负，所以状态的转移更新的方向 从x=m-1逆推
                     if (dp[x][y]>=&&dp[x+][y+sub[i]]<dp[x][y]+add[i]) {
                         dp[x+][y+sub[i]]=dp[x][y]+add[i];
                         //path[x+1][y+sub[i]] = i;    //这样定义是不行的  因为取m个   当前最优路径是动态的 中间结果可能被再次更新
                         path[x+][y+sub[i]] = path[x][y];
                         path[x+][y+sub[i]].push_back(i);
                     }
                 }
         }
         int ans; int a[];
         for (int i=;i<=fix;i++) if (dp[m][fix-i]>=||dp[m][fix+i]>=) {ans=i;break;}
         ans=dp[m][fix+ans]>dp[m][fix-ans]?ans:-ans;//判断ans的正负
         int sum1=(dp[m][fix+ans]+ans)/;
         int sum2=(dp[m][fix+ans]-ans)/;
         int v=fix+ans;
        /* for (int i=m;i>=1;i--) {
             a[i]=path[i][v];
             //cout<<v<<endl;
             v-=sub[a[i]];
         }*/
         printf( "Jury #%d\n", ++num );
         printf( "Best jury has value %d for prosecution and value %d for defence:\n", sum1,sum2);
        // for (int i=1;i<=m;i++)
             //cout<<" "<<a[i];
         for( int i=; i < m; i++ )
             printf( " %d", path[m][fix+ans][i]);
         cout<<"\n\n";
     }
     return ;
 }