TYVJ 1059 过河 解题报告

　　考的是动态规划的状态压缩吧，两个石子之间的距离若大于100就变成100，就压缩下来了。然后方程是f[i] = min{f[i - j]} + map[i];
　　代码：

#include <stdio.h>
#include <stdlib.h>
int map[10101];
int f[10101];
int stone[101];

int com(const void *a, const void *b)
{
	return *(int *)a - *(int *)b;
}

int main(int argc, char **argv)
{
	int i, j;
	int s, t, min;
	int n, m, base = 0;
	scanf("%d%d%d%d", &n, &s, &t, &m);
	for(i = 1; i <= m; i++){
		scanf("%d", &stone[i]);
	}
	if(s == t){
		for(i = 1; i <= m; i++){
			if(stone[i] % s == 0){
				base++;
			}
		}
		printf("%d\n", base);
		return 0;
	}

	qsort(&stone[1], m, sizeof(int), com);
	for(i = 1; i <= m; i++){
		j = stone[i];
		if(stone[i] - base > 100){
			stone[i] = stone[i - 1] + 100;
		}else{
			stone[i] = stone[i] - base + stone[i - 1];
		}
		base = j;
		map[stone[i]] = 1;
	}
	n = stone[m] + t;

	for(i = 1; i <= n; i++){
		f[i] = 0xFFFFFFF;
	}
	for(i = s; i <= n; i++){
		min = 0XFFFFFFF;
		for(j = s; j <= t; j++){
			if(min > f[i - j]){
				min = f[i - j];
			}
		}
		f[i] = min + map[i];
	}
	printf("%d\n", f[n]);
	return 0;
}