UVALive - 6892[区间DP]

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题目链接:https://vjudge.net/problem/UVALive-6892

解题思路:

一个表达式肯定可以分成左右两个部分然后以中间那个运算符合并之后的值，所以可以直接搞把数组复制两遍，然后搞个区间DP。

#include <bits/stdc++.h>
#define fi first 
#define se second
using namespace std;
typedef long long ll;
const int N = 1e3;
const int mx = 4e2 + 10;
int n,m,a[mx];
char c[mx];
ll g[mx][mx],f[mx][mx];
int main()
{
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++){
			scanf("%d ",a+i);
			scanf("%c",c+i);
			a[i+n] = a[i];
			c[i+n] = c[i];
		} 
		for(int i=1;i<=n;i++){
			for(int j=1;j<=2*n;j++){
				int l = j,r = i + j - 1;
				g[l][r] = -1e18,f[l][r] = 1e18;
				if(i==1) f[l][r] = g[l][r] = a[l];
				if(r>2*n) continue;
				for(int k=l;k<r;k++){
					if(c[k]=='-'||c[k]=='?'){
						f[l][r] = min(f[l][r],f[l][k]-g[k+1][r]);
						g[l][r] = max(g[l][r],g[l][k]-f[k+1][r]);
					}
					if(c[k]=='+'||c[k]=='?'){ 
						f[l][r] = min(f[l][r],f[l][k]+f[k+1][r]);
						g[l][r] = max(g[l][r],g[l][k]+g[k+1][r]);
					}
					if(c[k]=='*'||c[k]=='?'){
						f[l][r] = min(f[l][r],f[l][k]*f[k+1][r]);
						f[l][r] = min(f[l][r],f[l][k]*g[k+1][r]);
						f[l][r] = min(f[l][r],g[l][k]*f[k+1][r]);
						f[l][r] = min(f[l][r],g[l][k]*g[k+1][r]);
						g[l][r] = max(g[l][r],f[l][k]*f[k+1][r]);
						g[l][r] = max(g[l][r],f[l][k]*g[k+1][r]);
						g[l][r] = max(g[l][r],g[l][k]*f[k+1][r]);
						g[l][r] = max(g[l][r],g[l][k]*g[k+1][r]);
					}
				}
			}
		}
		for(int i=1;i<=n;i++)
		printf("%lld%lld",abs(f[i][i+n-1]),abs(g[i][i+n-1]));
		puts("");
	}
	return 0;
}