poj1015 Jury Compromise - dp，背包模型

以前年少无知，看到这题觉得无从入手，今天，终于看出来，这是一道背包题。

每个人，只有选和不选，如果选了，就会影响某个资源的值，而且人与人之间的顺序不影响答案，这种类型的题目，果断就是背包了。

 

#include <cstdio>
#include <vector>
#include <algorithm>
#include <cassert>
#include <cstdlib>
using namespace std;

int like[202];
int dislike[202];
int N, M;
int f[200 + 1][20 + 1][1600 + 2];
char pre[200 + 1][20 + 1][1600 + 2];
vector<int> ansList;
int testN;


const int nil = -2000000;
int dp(int n, int m, int d) {
  // printf("==== %d %d %d\n", n, m, d);
  return f[n][m][d + 800];
}



void input() {
  int i, j, k;
  scanf("%d %d", &N, &M);
  if (N == 0) exit(0);
  for (i = 1; i <= N; ++i) {
    scanf("%d %d", &(like[i]), &(dislike[i]));
  }


  for (i = 0; i < 201; ++i)
    for (j  = 0; j < 20 + 1; ++j)
      for (k = 0; k < 1600 + 2; ++k) {
        f[i][j][k] = nil;
        pre[i][j][k] = 'w';
      }
}


void traceBack(int n, int m, int d) {
  while (1) {
    char & son = pre[n][m][d + 800];
    // printf("*** %d %d %d %c\n", n, m, d, son);

    assert(n >= 1);

    if (son == '+') {
      ansList.push_back(n);
      d = d - (like[n] - dislike[n]);
      n = n - 1;
      m = m - 1;
      continue;
    }

    if (son == '-'){
      n = n - 1;
      continue;
    }

    if (son == 'e') break;
    if (son == 'E') {
      ansList.push_back(1);
      break;
    }

    assert(0);

  }
}



void run() {
  input();

  int n, m, d;
  for (n = 1; n <= N; ++n) {
    for (m = 0; m <= n && m <= M; ++m) {
      for (d = -800; d <= 800; ++d) {
          int & ans = f[n][m][d + 800];
          char & son = pre[n][m][d + 800];

          if (n == 1) {
            if (m == 0) {
              if (d == 0) { son = 'e'; ans = 0; continue; }
              else { son = 'x'; ans = -1; continue; }
            } else if (m == 1) {
              if (d == like[1] - dislike[1]) { son = 'E'; ans = like[1] + dislike[1]; continue; }
              else { son = 'y'; ans = -1; continue; }
            }

            assert(0);
          }

          ans = -1;
          son = 'p';
          int c = like[n] - dislike[n];
          if (n - 1 >= m && dp(n - 1, m, d) != -1) {
            ans = dp(n - 1, m, d);
            son = '-';
          }

          if (n >= m && m >= 1 && dp(n - 1, m - 1, d - c) != -1 && (ans == -1 || ans < dp(n - 1, m - 1, d - c) + like[n] + dislike[n]) ) {
            ans = dp(n - 1, m - 1, d - c) + like[n] + dislike[n];
            son = '+';
          }
      }
    }
  }


  int i, ans;
  for (i = 0; i <= 800; ++i) {
    if (dp(N, M, i) != -1 && dp(N, M, -i) == -1) {
      ans = i;
      break;
    }
    if (dp(N, M, i) == -1 && dp(N, M, -i) != -1) {
      ans = -i;
      break;
    }
    if (dp(N, M, i) != -1 && dp(N, M, -i) != -1 && dp(N, M, i) >= dp(N, M, -i)) {
      ans = i;
      break;
    }
    if (dp(N, M, i) != -1 && dp(N, M, -i) != -1 && dp(N, M, i) < dp(N, M, -i)) {
      ans = -i;
      break;
    }
  }



  ansList.clear();
  traceBack(N, M, ans);

  printf("Jury #%d\n", testN);
  printf("Best jury has value %d for prosecution and value %d for defence:\n",  (ans + dp(N, M, ans))/2,  -(ans - dp(N, M, ans))/2);

  sort(ansList.begin(), ansList.end());
  for (i = 0; i < ansList.size(); ++i) {
    printf(" %d", ansList[i]);
  }
  printf("\n\n");
}


int main() {
  for (testN = 1; true; ++testN)
    run();
  return 0;
}