DAG 模型

1.

#include<stdio.h>

#include<string.h>

#define MAXN 1010

int n, G[MAXN][MAXN];

int x[MAXN], y[MAXN], d[MAXN];

int dp(int i) {

  int& ans = d[i];

  if(ans > 0) return ans;

  ans = 1;

  for(int j = 1; j <= n; j++) if(G[i][j]) ans >?= dp(j)+1;

  return ans;

}
void print_ans(int i) {

  printf("%d ", i);

  for(int j = 1; j <= n; j++) if(G[i][j] && d[i] == d[j]+1) {

    print_ans(j);

    break;

  }

}
int main() {

  int i, j, ans, best;

  scanf("%d", &n);

  for(i = 1; i <= n; i++) {

    scanf("%d%d", &x[i], &y[i]);

    if(x[i] > y[i]) {

      int t = x[i]; x[i] = y[i]; y[i] = t;

    }

  }

  memset(G, 0, sizeof(G));

  for(i = 1; i <= n; i++)

    for(j = 1; j <= n; j++)

      if(x[i] < x[j] && y[i] < y[j]) G[i][j] = 1;
ans = 0;

  for(i = 1; i <= n; i++)

    if(dp(i) > ans) {

      best = i;

      ans = dp(i);

    }

  printf("%d\n", ans);

  print_ans(best);

  printf("\n");

  return 0;

}

2.

#include<stdio.h>

#include<string.h>

#define MAXN 1010

int n, G[MAXN][MAXN];

int x[MAXN], y[MAXN], d[MAXN];

int dp(int i) {

  int& ans = d[i];

  if(ans > 0) return ans;

  ans = 1;

  for(int j = 1; j <= n; j++) if(G[i][j]) ans >?= dp(j)+1;

  return ans;

}
int path[MAXN];

void print_all(int cur, int i) {

  path[cur] = i;

  if(d[i] == 1) {

    for(int j = 0; j <= cur; j++) printf("%d ", path[j]);

    printf("\n");

  }

  for(int j = 1; j <= n; j++) if(G[i][j] && d[i] == d[j]+1)

    print_all(cur+1, j);

}
int main() {

  int i, j, ans, best;

  scanf("%d", &n);

  for(i = 1; i <= n; i++) {

    scanf("%d%d", &x[i], &y[i]);

    if(x[i] > y[i]) {

      int t = x[i]; x[i] = y[i]; y[i] = t;

    }

  }

  memset(G, 0, sizeof(G));

  for(i = 1; i <= n; i++)

    for(j = 1; j <= n; j++)

      if(x[i] < x[j] && y[i] < y[j]) G[i][j] = 1;
ans = 0;

  for(i = 1; i <= n; i++) ans >?= dp(i);

  printf("%d\n", ans);
for(i = 1; i <= n; i++)

    if(ans ==  dp(i)) print_all(0, i);
return 0;

}

硬币问题（递推）
  coin.cpp(根据指标函数重建方案)
  coin2.cpp(递推时保存最优决策)

1.

#include<stdio.h>

#define MAXN 105

#define INF 1000000000

int V[MAXN], min[MAXN], max[MAXN];

int n, S;

void print_ans(int* d, int S) {

  for(int i = 1; i <= n; i++)

    if(S>=V[i] && d[S]==d[S-V[i]]+1) {

      printf("%d ", i);

      print_ans(d, S-V[i]);

      break;

    }

}
int main() {

  scanf("%d%d", &n, &S);

  for(int i = 1; i <= n; i++) scanf("%d", &V[i]);

  min[0] = max[0] = 0;

  for(int i = 1; i <= S; i++) {

    min[i] = INF;

    max[i] = -INF;

  }

  for(int i = 1; i <= S; i++)

    for(int j = 1; j <= n; j++)

      if(i >= V[j]) {

        min[i] <?= min[i-V[j]] + 1;

        max[i] >?= max[i-V[j]] + 1;

      }

  printf("%d %d\n", min[S], max[S]);

  print_ans(min, S);

  printf("\n");

  print_ans(max, S);

  printf("\n");

  return 0;

}

2.

#include<stdio.h>

#define MAXN 105

#define INF 1000000000

int V[MAXN], min[MAXN], max[MAXN], min_coin[MAXN], max_coin[MAXN];

int n, S;

void print_ans(int* d, int S) {

  while(S) {

    printf("%d ", d[S]);

    S -= V[d[S]];

  }

}
int main() {

  scanf("%d%d", &n, &S);

  for(int i = 1; i <= n; i++) scanf("%d", &V[i]);

  min[0] = max[0] = 0;

  for(int i = 1; i <= S; i++) {

    min[i] = INF;

    max[i] = -INF;

  }

  for(int i = 1; i <= S; i++)

    for(int j = 1; j <= n; j++)

      if(i >= V[j]) {

        if(min[i] > min[i-V[j]] + 1) {

          min[i] = min[i-V[j]] + 1;

          min_coin[i] = j;

        }

        if(max[i] < max[i-V[j]] + 1) {

          max[i] = max[i-V[j]] + 1;

          max_coin[i] = j;

        }

      }

  printf("%d %d\n", min[S], max[S]);

  print_ans(min_coin, S);

  printf("\n");

  print_ans(max_coin, S);

  printf("\n");

  return 0;

}

DAG 模型