题意：给定n个点的坐标，先问这些点是否能组成一个凸包，如果是凸包，问用不相交的线来切这个凸包使得凸包只由三角形组成，根据costi, j = |xi + xj| * |yi + yj| % p算切线的费用，问最少的切割费用。
思路：先找个凸包模板打上，然后区间DP即可
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define MAX 1000
#define INF 1000000000
#define min(a,b) ((a)<(b)?(a):(b))
struct point{
int x,y;
}p[MAX];
int cost[MAX][MAX],n,m;
int dp[MAX][MAX];
int abs(int x) {
return x < 0 ? -x : x;
}
point save[400],temp[400];
int xmult(point p1,point p2,point p0){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool cmp(const point& a,const point &b){
if(a.y == b.y)return a.x < b.x;
return a.y < b.y;
}
int Graham(point *p,int n) {
int i;
sort(p,p + n,cmp);
save[0] = p[0];
save[1] = p[1];
int top = 1;
for(i = 0;i < n; i++){
while(top && xmult(save[top],p[i],save[top-1]) >= 0)top--;
save[++top] = p[i];
}
int mid = top;
for(i = n - 2; i >= 0; i--){
while(top>mid&&xmult(save[top],p[i],save[top-1])>=0)top--;
save[++top]=p[i];
}
return top;
}
int Count(point a,point b) {
return (abs(a.x + b.x) * abs(a.y+b.y)) % m;
}
int main()
{
int i,j,k,r,u;
while (scanf("%d%d",&n,&m) != EOF) {
for (i = 0; i < n; ++i)
scanf("%d%d",&p[i].x,&p[i].y);
int tot = Graham(p,n); //求凸包
if (tot < n) printf("I can't cut.\n");
else {
memset(cost,0,sizeof(cost));
for (i = 0; i < n; ++i)
for (j = i + 2; j < n; ++j)
cost[i][j] = cost[j][i] = Count(save[i],save[j]);
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j)
dp[i][j] = INF;
dp[i][(i+1)%n] = 0;
}
for (i = n - 3; i >= 0; --i) //注意这三个for循环的顺序
for (j = i + 2; j < n; ++j) //因为要保证在算dp[i][j]时dp[i][k]和dp[k][j]时已经计算，所以i为逆序，j要升序
for (k = i + 1; k <= j - 1; ++k)
dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
printf("%d\n",dp[0][n-1]);
}
}
}
Description
You want to hold a party. Here's a polygon-shaped cake on the table. You'd like to cut the cake into several triangle-shaped parts for the invited comers. You have a knife to cut. The trace of each cut is a line segment, whose two endpoints are two vertices of the polygon. Within the polygon, any two cuts ought to be disjoint. Of course, the situation that only the endpoints of two segments intersect is allowed.
The cake's considered as a coordinate system. You have known the coordinates of vexteces. Each cut has a cost related to the coordinate of the vertex, whose formula is costi, j = |xi + xj| * |yi + yj| % p. You want to calculate the minimum cost.
NOTICE: input assures that NO three adjacent vertices on the polygon-shaped cake are in a line. And the cake is not always a convex.
Input
There're multiple cases. There's a blank line between two cases. The first line of each case contains two integers, N and p (3 ≤ N, p ≤ 300), indicating the number of vertices. Each line of the following N lines contains two integers, x and y (-10000 ≤ x, y
Output
If the cake is not convex polygon-shaped, output "I can't cut.". Otherwise, output the minimum cost.
Sample Input
3 3 0 0 1 1 0 2
Sample Output
0
