【LOJ】#2349. 「JOI 2017/2018 决赛」团子制作

题解

有意思的一个dp，我们对G计数，发现如果不在同一条对角线上的G肯定不会互相影响，所以我们对于每一条对角线dp
dp的方式是枚举这个G以什么方式放，横着还是竖着，还是不放

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
#include <queue>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 100005
#define MAXM 200005
#define eps 1e-3
#define RG register
#define calc(x) __builtin_popcount(x)
#define pLI pair<int64,int>
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M,ans,dp[3][3005];
char s[3005][3005];
bool vis[3005][3005],has[3005][3005];
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) scanf("%s",s[i] + 1);
    for(int h = 2 ; h <= N + M ; ++h) {
	memset(dp,0,sizeof(dp));
	int tmp = 0;
	for(int i = max(1,h - M) ; i <= min(h - 1,N) ; ++i) {
	    int j = h - i;
	    dp[0][i] = max(max(dp[0][i - 1],dp[1][i - 1]),dp[2][i - 1]);
	    if(s[i][j] == 'G') {
		if(s[i - 1][j] == 'R' && s[i + 1][j] == 'W') {
		    dp[1][i] = max(dp[0][i - 1],dp[1][i - 1]) + 1;
		}
		if(s[i][j - 1] == 'R' && s[i][j + 1] == 'W') {
		    dp[2][i] = max(dp[0][i - 1],dp[2][i - 1]) + 1;
		}
	    }
	    tmp = max(tmp,max(max(dp[0][i],dp[1][i]),dp[2][i]));
	}
	ans += tmp;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}