Algorithm: POJ 3207 ikki ' s Story IV

"The main effect of the topic"

There are n numbers on a circle: 0,1,2 ... N-1. Then there is the M line 22 connecting these dots, which can be inside the circle or outside the circle. Ask if there is a way to make these lines do not intersect.

Ideas

In lines, each line is either inside the circle or outside the circle, a typical 2-sat model.

Code

#include <iostream> #include <cstdio> #include <cstring> using namespace std;  
    
typedef long long Int64;  
const int MAXN = 1010;  
const int VN = MAXN*2;  
const int EN = VN*VN/2;  
int n, m;  
    
int arr[vn][2];  
    struct graph{int size;  
    int HEAD[VN];  
    struct{int V, next;  
    
    }e[en];  
        void init () {size = 0;  
    Memset (Head,-1, sizeof (head));  
        } void Addedge (int u, int v) {e[size].v = v;  
        E[size].next = Head[u];  
    Head[u] = size++;  
    
}}g;  
        Class two_sat{Public:bool Check (const graph& g, const int n) {SCC (g, N);  
        for (int i=0; i<n; ++i) if (belong[i] = = Belong[i+n]) return false;  
    return true;  
        } private:void Tarjan (const graph& g, const int u) {int V;  
        Dfn[u] = low[u] = ++idx;  
        sta[top++] = u; Instack[u] = True;  
            for (int e=g.head[u]; E!=-1 e=g.e[e].next) {v = g.e[e].v;  
                if (dfn[v] = =-1) {Tarjan (g, v);  
            Low[u] = min (Low[u], low[v]);  
            }else if (Instack[v]) {Low[u] = min (Low[u], dfn[v]);  
            } if (dfn[u] = = Low[u]) {++bcnt;  
                do{v = sta[--top];  
                INSTACK[V] = false;  
            BELONG[V] = bcnt;  
        }while (U!= v);  
        } void SCC (const graph& g, const int n) {idx = top = bcnt = 0;  
        memset (DFN,-1, sizeof (DFN));  
        memset (instack, 0, sizeof (instack));  
    for (int i=0; i<2*n; ++i) if (dfn[i] = = 1) Tarjan (g, I);  
    } private:int idx, top, bcnt;  
    int DFN[VN];  
    int LOW[VN];  
    int STA[VN];  
    int BELONG[VN];  
    
BOOL INSTACK[VN];  
    
    
}sat; boolIscross (int* A, int *b) {return a[0]>b[0]&&a[0]<b[1] && (a[1]<b[0) | | A[1]>B[1]) | | A[1]>b[0]&&a[1]<b[1] && (a[0]<b[0) | |  
A[0]>B[1]);  
        int main () {while (~scanf ("%d%d", &n, &m)) {g.init ();  
            for (int i=0; i<n; ++i) {scanf ("%d%d", &arr[i][0], &arr[i][1]);  
        if (Arr[i][0] > Arr[i][1]) swap (arr[i][0], arr[i][1]); for (int i=0; i<n; ++i) {for (int j=i+1; j<n; ++j) {if (Iscross (arr[i)  
                    , Arr[j]) {G.addedge (I, j+n);  
                    G.addedge (I+n, J);  
                    G.addedge (J, I+n);  
                G.addedge (J+n, i);  
        }} if (Sat.check (g, N)) puts ("Panda is telling the truth ...");  
    
    Else puts ("The evil Panda is lying again");  
return 0; }

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/