Dynamic Programming Solution to Activity Selection Problem

#include <stdio.h>#define MAX_ACTIVITY_NUM100struct activity{int start_time;int end_time;};static struct activity g_activities[MAX_ACTIVITY_NUM];static int g_solution[MAX_ACTIVITY_NUM][MAX_ACTIVITY_NUM];static int g_count;static int g_temp[MAX_ACTIVITY_NUM][MAX_ACTIVITY_NUM];int find_partition(struct activity *pActivities, int p, int q){int data = pActivities[q].end_time;int i = p-1, j;for(j = p; j < q; j++){if(pActivities[j].end_time < data){i++;struct activity tmp = pActivities[j];pActivities[j] = pActivities[i];pActivities[i] = tmp;}}i++;struct activity tmp = pActivities[q];pActivities[q] = pActivities[i];pActivities[i] = tmp;return i;}void quick_sort(struct activity *pActivities, int i, int j){if(i < j){int p = find_partition(pActivities, i, j);quick_sort(pActivities, i, p -1);quick_sort(pActivities, p+1, j);}}void find_optimal_activities(struct activity *pActivities, int start, int end){int step, i, j, k;for(step = 2; step < end -start + 1; step++){for(i = start; i < end - step + 1; i++){j = i + step;if(pActivities[i].end_time > pActivities[j].start_time){continue;}int temp = 0;int k_max;for(k = i + 1; k < j; k++){if( pActivities[k].start_time >= pActivities[i].end_time &&pActivities[k].end_time <= pActivities[j].start_time){if( temp <  g_temp[i][k] + g_temp[k][j] + 1){temp = g_temp[i][k] + g_temp[k][j] + 1;k_max = k;}}}if(temp > g_temp[i][j] ){g_temp[i][j] = temp;g_solution[i][j] = k_max;}}}}void print_solution(int i, int j){if( g_solution[i][j] != 0){print_solution(i, g_solution[i][j]);printf("[%d, %d]: %d\n", i, j, g_solution[i][j]);print_solution(g_solution[i][j], j);}}int main(void){freopen("data.txt", "r", stdin);int count;scanf("%d", &count);int p;for(p = 0; p < count; p++){g_activities[0].start_time = g_activities[0].end_time = -1 << 31;scanf("%d", &g_count);int i;for(i = 1; i <= g_count; i++){scanf("%d %d", &g_activities[i].start_time, &g_activities[i].end_time);}g_activities[g_count].start_time = g_activities[g_count].end_time = (1 << 31) -1 ;quick_sort(g_activities, 0, g_count + 1);int j;for( i = 0; i < MAX_ACTIVITY_NUM; i++){for(j = 0; j < MAX_ACTIVITY_NUM; j++){g_temp[i][j] = 0;g_solution[i][j] = 0;}}find_optimal_activities(g_activities, 0, g_count + 1);printf("max count of compatible activities: %d\n", g_temp[0][g_count + 1]);print_solution(0, g_count + 1);}return 0;}

参考：

http://blog.csdn.net/cyp331203/article/details/43242209