uva1629
題目描述：有一個n行m列（1≤n，m≤20）的網格蛋糕上有一些櫻桃。每次可以用一刀沿著網格線把 蛋糕切成兩塊，并且只能夠直切不能拐彎。要求最后每一塊蛋糕上恰好有一個櫻桃，且切割 線總長度最小。
分析：dp切割行和列
#include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<stdio.h> #include<string> #include<vector> #include<queue> #include<map> #include<sstream> #include<cassert> #include<set> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; const int maxn = 25; int n, m; int dp[maxn][maxn][maxn][maxn]; int g[maxn][maxn]; int judge(int u, int d, int l, int r) {int ok = 0;for (int i = d; i <= u; i++) {for (int j = l; j <= r; j++) {if (g[i][j]) {if (ok)return 2;else ok = 1;}}}return ok; }int solve(int u, int d, int l, int r) {int &k = dp[u][d][l][r];if (k != -1)return k;int t = judge(u, d, l, r);if (t == 1)return 0;else if (t == 0)return INF;k = INF;for (int i = l; i < r; i++) {k = min(k, solve(u, d, l, i) + solve(u, d, i + 1, r) + u - d + 1);}for (int i = d; i < u; i++) {k = min(k, solve(i, d, l, r) + solve(u, i+1, l, r) + r - l + 1);}return k; } int main() {int k, t = 0;while (cin >> n >> m >> k) {memset(g, 0, sizeof(g));memset(dp, -1, sizeof(dp));int u, v;while (k--) {cin >> u >> v;g[u][v] = 1;}cout << "Case " << ++t << ": " << solve(n, 1, 1, m) << endl;;}return 0; }?
總結
