洛谷 P4342 [IOI1998]Polygon

题目传送门

解题思路:

一道环形dp,只不过有个地方要注意,因为有乘法,两个负数相乘是正数,所以最小的数是负数,乘起来可能比最大值大,所以要记录最小值(这道题是紫题的原因).

AC代码:

#include<cstdio>
#include<iostream>
#include<cstring> 

using namespace std;

int n,a[110],maxf[110][110],minf[110][110],ans = -2147418364; 
char c[110];

int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n; i++) {
        cin >> c[i] >> a[i];//char和string最好用cin,因为用了scanf调了好长时间 
        a[i+n] = a[i];
        c[i+n] = c[i];
    }
    memset(minf,0x80,sizeof(minf));//最小值 
    memset(maxf,0x3f,sizeof(maxf));//最大值 
    for(int i = 1;i <= n + n - 1; i++) {
        maxf[i][i] = a[i];
        minf[i][i] = a[i];
    }    
    for(int p = 2;p <= n; p++)
        for(int i = 1;i + p - 1 <= n + n - 1; i++){ 
            int j = i + p - 1; 
            for(int k = i + 1;k <= j; k++)
                if(c[k] == 'x') {//如果是乘 
                    minf[i][j] = max(minf[i][j],max(minf[i][k-1] * minf[k][j],max(maxf[i][k-1] * minf[k][j],max(minf[i][k-1] * maxf[k][j],maxf[i][k-1] * maxf[k][j]))));
                    maxf[i][j] = min(maxf[i][j],min(minf[i][k-1] * minf[k][j],min(maxf[i][k-1] * minf[k][j],min(minf[i][k-1] * maxf[k][j],maxf[i][k-1] * maxf[k][j]))));
                }
                else {
                    minf[i][j] = max(minf[i][j],minf[i][k-1] + minf[k][j]);
                    maxf[i][j] = min(maxf[i][j],maxf[i][k-1] + maxf[k][j]);
                }
        } 
    for(int i = 1;i <= n; i++)
        ans = max(ans,minf[i][i+n-1]);
    printf("%d\n",ans);
    for(int i = 1;i <= n; i++)
        if(minf[i][i+n-1] == ans)
            printf("%d ",i);
    return 0;
}