TYVJ 1124 花店橱窗 解题报告

　　f[i][j]代表将第i束花插进第j个花屏所能获得的最大的分，方程是: f[i][j] = max(f[i - 1][k] + map[i][j]) (i -1=<k<j)，代码如下：

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

void output(int a, int b)
{
	if(a == 1){
		printf("%d", b);
		return;
	}
	output(a - 1, p[a][b]);
	printf(" %d", b);
}

int main(int argc, char **argv)
{
	int i, j, k;
	int m, n;
	scanf("%d%d", &n, &m);
	for(i = 1; i <= n; i++){
		for(j = 1; j <= m; j++){
			scanf("%d", &map[i][j]);
		}
	}
	for(i = 1; i <= n; i++){
		for(j = i; j <= m - n + i; j++){
			f[i][j] = -100000000;
			for(k = i - 1; k < j; k++){
				if(f[i - 1][k] + map[i][j] > f[i][j]){
					f[i][j] = f[i - 1][k] + map[i][j];
					p[i][j] = k;
				}
			}
		}
	}
	for(i = n, k = 0; i <= m; i++){
		if(k < f[n][i]){
			k = f[n][i];
			j = i;
		}
	}
	printf("%d\n", f[n][j]);
	output(n, j);
	printf("\n");
	return 0;
}