Algorithm: UVa 1146

"The main effect of the topic"

There are n planes to land, and each plane can opt for an "early landing" or "late landing", which cannot be landed at any other time. Give each aircraft "early landing" and "play landing" time, asked how to arrange the landing, can make any two aircraft of the land time interval is minimal, output minimum.

Ideas

Water problem, the minimum interval of two minutes, with 2-sat judgment.

Code

#include <iostream> #include <queue> #include <cstdio> #include <cstring> #include <CMATH&G  
T  
    
using namespace Std;  
const int MAXN = 2005;  
const int VN = MAXN*2;  
const int EN = MAXN*MAXN*4;  
int n, m;  
    
int t[maxn][2];  
    struct graph{int size, 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 () {SCC ();  
        for (int i=0; i<n; ++i) if (belong[i] = = Belong[i+n]) return false;  
    return true;  
        } private:void Tarjan (const int u) {int V;  
        Low[u] = dfn[u] = ++idx;  
        Instack[u] = true;  
        sta[top++] = u;  
            for (int e=g.head[u]; E!=-1 e=g.e[e].next) {v = g.e[e].v; IF (dfn[v] = =-1) {Tarjan (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 () {idx = top = bcnt = 0;  
        memset (instack, 0, sizeof (instack));  
        memset (DFN,-1, sizeof (DFN));  
    for (int i=0; i<2*n; ++i) if (dfn[i] = = 1) tarjan (i);  
    } private:int idx, top, bcnt;  
    int STA[VN], low[vn], DFN[VN], BELONG[VN];  
BOOL INSTACK[VN];  
    
    
}sat;  
    void buildgraph (int minT) {g.init ();  
            for (int i=0; i<n; ++i) {for (int j=i+1; j<n; ++j) {int ta1=t[i][0], ta2=t[i][1];  
   int tb1=t[j][0], tb2=t[j][1]; 
            if (ABS (TA1-TB1) < MinT) {G.addedge (I, j+n);  
            G.addedge (J, I+n);  
                } if (ABS (TA1-TB2) < MinT) {G.addedge (i, j);  
            G.addedge (J+n, i+n);  
                } if (ABS (TA2-TB1) < MinT) {G.addedge (i+n, j+n);  
            G.addedge (J, I);  
                } if (ABS (TA2-TB2) < MinT) {G.addedge (i+n, J);  
            G.addedge (J+n, i); int main () {while (~SCANF ("%d", &n)) {for (int i=0; i <n;  
    
        ++i) scanf ("%d%d", &t[i][0), &t[i][1]);  
        int l=0, R=1e7, Mid, ans=-1;  
            while (L < r) {mid = (l+r) >>1;  
            Buildgraph (mid);  
                if (Sat.check ()) {ans = mid;  
            L = mid+1;  
        }else r = Mid; }  
        printf ("%d\n", ans);  
return 0; }

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