uva 10558(dp)

来源：互联网发布：天下三铭尘父子知乎编辑：程序博客网时间：2024/05/16 08:41

题意：一个100*100矩阵上有n个点，给出n个点的坐标，然后划了m条竖线，和m条竖线的坐标，然后问如果要切A条线，必须切1和100这两条线，问怎样切横线可以让区域最多，区域必须是有点存在的，边界上的点属于点右上区域。

题解：想不出来，参考网上的人的题解，通过初始化用了个数组s[i][j]表示从第i条横线到第j条横线有多少个区域，然后就可以普通的dp，f[i][j]表示从第i条线到最后且还剩j条横线需要切割最多切多少个区域出来，状态转移方程：f[i][j] = max{f]i][j]，f[k][j - 1] + s[i][k]}。然后用path[i][j]存剩余100-i条横线剩余j次切割要切在哪条线上，根据path[i][j]递归输出路径。

#include <stdio.h>#include <string.h>const int N = 105;int n, m, A;int c[N], g[N][N], exi[N][N], d[N][N], s[N][N], f[N][N], path[N][N];void init() {for (int i = 1; i < m; i++) {for (int j = 1; j < 100; j++) {exi[j][i] = d[j][i] = 0;for (int k = c[i]; k < c[i + 1]; k++)if (g[j][k]) {exi[j][i] = 1;break;}d[j][i] = exi[j][i] + d[j - 1][i];}}for (int i = 1; i < 100; i++) {for (int j = i + 1; j <= 100; j++) {s[i][j] = 0;for (int k = 1; k < m; k++)if (d[j - 1][k] - d[i][k] + exi[i][k])s[i][j]++;}}}int dp(int cur, int p) {if (f[cur][p] != -1)return f[cur][p];if (p == 0)return f[cur][p] = s[cur][100];f[cur][p] = 0;for (int i = cur + 1; i < 100; i++) {if (p <= 100 - i) {int temp = dp(i, p - 1);if (f[cur][p] < temp + s[cur][i]) {f[cur][p] = temp + s[cur][i];path[cur][p] = i;}}}return f[cur][p];}void print_path(int cur, int p) {if (p == 0)return;printf(" %d", path[cur][p]);print_path(path[cur][p], p - 1);}int main() {while (scanf("%d", &n) && n != -1) {memset(f, -1, sizeof(f));memset(g, 0, sizeof(g));int a, b;for (int i = 0; i < n; i++) {scanf("%d%d", &a, &b);g[b][a] = 1;}scanf("%d", &m);for (int i = 1; i <= m; i++)scanf("%d", &c[i]);scanf("%d", &A);init();dp(1, A - 2);printf("%d 1", A);print_path(1, A - 2);printf(" 100\n");}return 0;}

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