题目意思：给定N对数，取期中的m对，要求取出sigma（a）与sigma(b)的差最小，如果有多种，输出两者和最大的一种。。。
DP求解.f[i,j,k]表示前i对数中选j对,第一个数之和减第二个数之和的差为k,这种情况下第二个数之和的最大值.
显然,f[i,j,k]=max(f[i-1,j,k],f[i-1,j-1,k-d[i]+d[i]]+p[i])；；
View Code
1 /* 2 State:Accept 3 Time:2013.2.28 4 */ 5 6 #include <iostream> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <fstream> 10 #include <algorithm> 11 #include <cstring> 12 #include <string> 13 #include <cmath> 14 #define minn -10000; 15 using namespace std; 16 int f[205][25][805] , finalx, finaly , n , m; 17 int result[25] , d[202], p[202] , ansx, ansy , ans ,test; 18 19 void init(){ 20 for (int i = 1; i <= n; ++i) 21 scanf("%d%d",&d[i],&p[i]); 22 memset(result, 0, sizeof(result)); 23 memset(f , 0 ,sizeof(f)); 24 } 25 26 27 void dp(){ 28 int k2 ; 29 for (int i = 0; i <= n; ++i) 30 for (int j = 0; j <= m; ++j) 31 for (int k = 0; k <= 800; ++k) 32 if (j == 0 && k == 400) 33 f[i][j][k] = 0; 34 else f[i][j][k] = minn; 35 36 37 for (int i = 1; i <= n; ++i) 38 for (int j = 1; j <= min(m,i); ++j) 39 for (int k = 0; k <= 800; ++k){ 40 k2 = k - d[i] + p[i]; 41 if (k2 < 0 || k2 > 800) continue; 42 f[i][j][k] = max(f[i - 1][j - 1][k2] + d[i] + p[i] , f[i - 1][j][k]); 43 } 44 printf("Jury #%d\n", test); 45 } 46 47 void find(int i , int j , int k){ 48 if (k == 400 && j == 0) return; 49 if (f[i][j][k] == f[i-1][j][k]) find(i - 1, j ,k); 50 else { 51 result[j] = i; 52 ansx += p[i]; 53 ansy += d[i]; 54 find(i - 1, j - 1 , k + p[i] - d[i]); 55 56 } 57 } 58 59 void printans(){ 60 bool bo = false; 61 int temp; 62 ans = minn; 63 ansx = ansy = 0; 64 for (int k = 0; k <= 400; ++k) if (ans < 0) 65 for (int i = 1; i <= n; ++i) 66 if (f[i][m][400 - k] >= 0 || f[i][m][400 + k] >= 0 ){ 67 temp = max(f[i][m][400 - k] , f[i][m][400 + k]); 68 if (temp <= ans) continue; 69 ans = temp; 70 finalx = i; 71 finaly = 400 + k; 72 if (f[i][m][400 - k] > f[i][m][400 + k]) 73 finaly = 400 -k; 74 } 75 find(finalx , m , finaly); 76 printf("Best jury has value %d for prosecution and value %d for defence:\n",ansy, ansx); 77 for (int i = 1 ; i <= m; ++i) 78 printf("%d ",result[i]); 79 printf("\n\n"); 80 } 81 82 int main(){ 83 freopen("poj1015.in","r",stdin); 84 freopen("poj1015.out","w",stdout); 85 scanf("%d%d",&n , &m); 86 while (n && m ){ 87 ++ test; 88 init(); 89 dp(); 90 printans(); 91 scanf("%d%d",&n , &m); 92 93 } 94 fclose(stdin); 95 fclose(stdout); 96 }
注意最后输出，有点麻换。。