【动态规划】[UVA 437]The Tower of Babylon

令dp(i,j)表示用第i个方块做底其中第j条边做高的时候所可以得到的最大高度那么显然dp(i,j)=min{dp(k,p)}+h(i,j)同时要求第k个的底面严格小于i的底面。这道题就是这样。

#include 
#include 
#include 
using namespace std;
const int MAXN = 1000;
int f[MAXN+10][3];
int d[MAXN+10][3];
bool check(int i, int h1, int j, int h2){
    int a1=d[i][0], b1=d[i][1], a2=d[j][0], b2=d[j][1];
    if(h1 != 2){
        if(h1 == 1)
            b1 = d[i][2];
        else a1 = d[i][2];
    }
    if(h2 != 2){
        if(h2 == 1)
            b2 = d[j][2];
        else a2 = d[j][2];
    }
    if(a1 > b1) swap(a1, b1);
    if(a2 > b2) swap(a2, b2);
    if(a1 < a2 && b1 < b2) return true;
    return false;
}
int rans, n;
int dp(int u, int m){
    int &ans = f[u][m];
    if(ans > 0) return ans;
    for(int i=1;i<=n;i++){
        for(int v=0;v<3;v++)
            if(check(i, v, u, m))
                ans = max(ans, dp(i, v));
    }
    ans += d[u][m];
    return rans=max(rans,ans), ans;
}
int main(){
    int ccnt = 0;
    while(scanf("%d", &n)!=EOF && n!=0){
        memset(f, 0, sizeof f); rans = 0;
    for(int i=1;i<=n;i++)
        scanf("%d%d%d", &d[i][0], &d[i][1], &d[i][2]);
    for(int i=1;i<=n;i++)
        for(int v=0;v<3;v++)
            dp(i, v);
    printf("Case %d: maximum height = %d\n", ++ccnt, rans);
}

    return 0;
}

转载于:https://www.cnblogs.com/JeremyGJY/p/5921709.html