经典动态规划题目(POJ 1015)

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在文章开始时，先声明一下，小弟刚开始接触acm，水平有限，所以这里给出的程序是阉割版的，只实现了关键功能，无法ac

这道题类似于经典的数组分割问题，可参见《编程之美》2.18

#include <stdio.h>#include <memory.h>const int MAX_N = 202;const int MAX_M = 22;const int MAX_SUM = 1001;bool dp[MAX_N][MAX_SUM];int path[MAX_N][MAX_SUM];bool is_neg[MAX_N][MAX_SUM];int pcs[MAX_N];int dfs[MAX_N];int min2(int a, int b){return a < b ? a : b;}void solve(int n, int m){int i, j, k;int base = m*20;dp[0][base] = true;for (k = 1; k <= n; ++k){for (i = min2(k,m); i >= 1; --i){for (j = 0; j <= 2*base; ++j){int val = dfs[k-1] - pcs[k-1];if (j >= val && j - val <= 2*base && dp[i-1][j-val]){dp[i][j] = true;path[i][j] = k;}}}}i = m;j = base;while (j <= 2*base && !dp[i][j]){j++;}k = base - 1;while (k >= 0 && !dp[i][k]){k--;}printf("%d %d\n", j-base, k-base);k=j;while (i >= 1){printf("%d\n", path[i][k]);k -= (dfs[path[i][k]-1] - pcs[path[i][k]-1]);i--;}}void input(){int n, m;while (scanf("%d %d", &n, &m) && n != 0 && m != 0){int i;for (i = 0; i < n; ++i){scanf("%d %d", &pcs[i], &dfs[i]);}memset(dp, 0, sizeof(dp));solve(n, m);}}int main(){input();return 0;}