DP紫书汇总

这是按照专题对一些题目的汇总，总体上是按照lrj的《算法艺术与新自学竞赛》的介绍进行的总结
其实就是贴的lrj的题解+部分自己的理解……

DAG上的动态规划

A Spy in the Metro (UVA - 1025)

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n, T, t[500], m, d, dp[500][500], Case;bool has_train[500][500][2];int main() {    while (scanf("%d", &n) != EOF && n) {        scanf("%d", &T);        for (int i = 1; i < n; ++i) scanf("%d", &t[i]), dp[T][i] = 1 << 30;        dp[T][n] = 0;        scanf("%d", &m);        for (int i = 1; i <= m; ++i) {            scanf("%d", &d);            int cnt = 0;            for (int j = 1; j <= n; ++j) has_train[d + cnt][j][0] = 1, cnt += t[j];        }        scanf("%d", &m);        for (int i = 1; i <= m; ++i) {            scanf("%d", &d);            int cnt = 0;            for (int j = n; j >= 1; --j) has_train[d + cnt][j][1] = 1, cnt += t[j - 1];        }        for (int i = T - 1; i >= 0; i--)            for (int j = 1; j <= n; j++) {                dp[i][j] = dp[i + 1][j] + 1;                if (j < n && has_train[i][j][0] && i + t[j] <= T)                    dp[i][j] = min(dp[i][j], dp[i + t[j]][j + 1]);                if (j > 1 && has_train[i][j][1] && i + t[j - 1] <= T)                    dp[i][j] = min(dp[i][j], dp[i + t[j - 1]][j - 1]);            }        printf("Case Number %d: ", ++Case);        if (dp[0][1] >= 1 << 30) printf("impossible\n");        else printf("%d\n", dp[0][1]);        memset(has_train, 0, sizeof(has_train));        memset(t, 0, sizeof(t));        memset(dp, 0, sizeof(dp));     }    return 0;}

The Tower of Babylon (UVA - 437)

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, a[50][3], f[50][3], Case;inline int dp(int x, int y) {    int& ans = f[x][y];    if (ans > 0) return ans;    ans = 0;    int cnt = 0, v1[2], v2[2];    for (int i = 0; i < 3; ++i) if (i != y) v1[cnt++] = a[x][i];    for (int i = 1; i <= n; ++i)        for (int j = 0; j < 3; ++j) {            cnt = 0;            for (int k = 0; k < 3; ++k) if (k != j) v2[cnt++] = a[i][k];            if (v2[0] < v1[0] && v2[1] < v1[1]) ans = max(ans, dp(i, j));//判断是否严格的小于下方的立方体底面的长宽        }    ans += a[x][y];    return ans;}int main() {    while (scanf("%d", &n) == 1 && n) {        for (int i = 1; i <= n; ++i) {            scanf("%d %d %d", &a[i][0], &a[i][1], &a[i][2]);            sort(a[i], a[i] + 3);        }        memset(f, 0, sizeof(f));        int res = 0;        for (int i = 1; i <= n; ++i)            for (int j = 0; j < 3; ++j)                res = max(res, dp(i, j));        printf("Case %d: maximum height = %d\n", ++Case, res);        memset(a, 0, sizeof(a));    }    return 0;}