ZOJ3537 Cake 题意：

给你几何形状的蛋糕，你需要按照顶点来切蛋糕，每次切蛋糕会产生一个贡献，\(val=cost_{i, j} *|x_i + x_j| * |y_i + y_j| % p\) 问你把蛋糕切成几个三角形所需要的最小花费

题解：

区间dp好题

首先我们维护 一个凸包，如果这个几何形状只有三个点，那么就不需要切，花费为0

然后维护出一个凸包出来，如果满足凸包的条件的话，我们就开始切三角形

很显然我们需要维护一个花费最小的三角形剖分

设置dp[i][j]为从顶点i到顶点j所围成凸多边形的最优解。

枚举切点k (i < k < j)

\(dp[i][j] = min(dp[i][k] + dp[k][j] + cost[i][k] + cost[k][j])\)

代码： 
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
LL quick_pow(LL x, LL y) {
    LL ans = 1;
    while(y) {
        if(y & 1) {
            ans = ans * x % mod;
        } x = x * x % mod;
        y >>= 1;
    } return ans;
}
struct Point {
    int x, y;
    Point() {}
    Point(int xx, int yy): x(xx), y(yy) {}
    void read() {
        scanf("%d%d", &x, &y);
    }
    bool operator < (const Point &other)const {
        if(x == other.x)
            return y < other.y;
        return x < other.x;
    }
    Point operator - (const Point &other)const {
        return Point(x - other.x, y - other.y);
    }
};
Point P[400], ch[400];
int n, m;
double Cross(Point A, Point B) {
    return A.x * B.y - A.y * B.x;
}
int ConvexHull() {
    sort(P, P + n);
    int cnt = 0;
    for(int i = 0; i < n; i++) {
        while(cnt > 1 && Cross(ch[cnt - 1] - ch[cnt - 2], P[i] - ch[cnt - 2]) <= 0) cnt--;
        ch[cnt++] = P[i];
    }
    int k = cnt;
    for(int i = n - 2; i >= 0; i--) {
        while(cnt > k && Cross(ch[cnt - 1] - ch[cnt - 2], P[i] - ch[cnt - 2]) <= 0) cnt--;
        ch[cnt++] = P[i];
    }
    if(n > 1) cnt--;
    return cnt;
}
int calc(Point a, Point b) {
    return (abs(a.x + b.x) * abs(a.y + b.y)) % m;
}
int f[505][505];
int dp[505][505];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    while(~scanf("%d%d", &n, &m)) {
        for(int i = 0; i < n; i++) {
            scanf("%d%d", &P[i].x, &P[i].y);
        }
        if(n == 3) {
            puts("0");
            continue;
        }
        if(ConvexHull() < n) {
            printf("I can't cut.\n");
        } else {
            for(int i = 0; i < n; i++) {
                for(int j = i + 2; j < n; j++) {
                    f[i][j] = f[j][i] = calc(ch[i], ch[j]);
                }
            }
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    dp[i][j] = INF;
                }
                dp[i][(i + 1) % n] = 0;
            }
            for(int i = n - 3; i >= 0; i--) {
                for(int j = i + 2; j < n; j++) {
                    for(int k = i + 1; k < j; k++) {
                        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + f[i][k] + f[k][j]);
                    }
                }
            }
            printf("%d\n", dp[0][n - 1]);
        }
    }
    return 0;
}
每一个不曾刷题的日子 都是对生命的辜负 从弱小到强大，需要一段时间的沉淀，就是现在了 ~buerdepepeqi