LightOJ 1004 - Monkey Banana Problem （dp）

题意: 变形数塔
分析: 水题 - - 注意边界, 然后开大数组可以无视
代码:

////  Created by TaoSama on 2015-10-27//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, a[205][205], dp[205][205];void getMax(int &x, int y) {x = max(x, y);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        memset(a, 0, sizeof a);        for(int i = 1, cnt = 0; i < n << 1; ++i) {            cnt += i <= n ? 1 : -1;            for(int j = 1; j <= cnt; ++j)                scanf("%d", &a[i][j]);        }        memset(dp, 0, sizeof dp);        for(int i = 1, cnt = 1; i < n << 1; ++i) {            int x = i <= n ? 1 : -1;            cnt += x;            for(int j = 1; j <= cnt; ++j) {                getMax(dp[i][j], dp[i - 1][j] + a[i][j]);                getMax(dp[i][j], dp[i - 1][j - x] + a[i][j]);            }        }        printf("Case %d: %d\n", ++kase, dp[(n << 1) - 1][1]);    }    return 0;}