Hdu 2236 Untitled II maximum match + two points

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Hdu 2236

Problem Solving Ideas:

The rows and columns are interpreted as the ends of the two-point graph, and the given matrix is all the edges in the two-minute graph.

If the binary graph can match exactly, then the n element interval of the different columns is (Min_edge,max_edge), which refers to those edges that make up the exact match

The problem needs to solve the minimum interval length

We can divide the interval length (0~100), the lower bound of each enumeration interval

The last Maxl is the answer.

Code:

#include <iostream> #include <cstdio> #include <cstring> #define MAXN 105using namespace Std;int map[    Maxn][maxn];int link[maxn];int vis[maxn];int n,minv,maxv;int maxl,minl,l,mid;int dfs (int u) {for (int i=1;i<=n;i++)            {if (!vis[i]&&map[u][i]>=l&&map[u][i]<=l+mid) {vis[i]=1; if (link[i]==-1| |                DFS (Link[i])) {link[i]=u;            return 1; }}} return 0;}    int hungry () {memset (link) (link,-1,sizeof);        for (int i=1;i<=n;i++) {memset (vis,0,sizeof (VIS));    if (!dfs (i)) return false; } return true;    int main () {int T;    scanf ("%d", &t);        while (t--) {minv=105,maxv=0;        scanf ("%d", &n);                for (int. I=1; i<=n; i++) for (int j=1; j<=n; J + +) {scanf ("%d", &map[i][j]);                if (MAP[I][J]&GT;MAXV) maxv=map[i][j]; if (MAP[I][J]&LT;MINV) minV=MAP[I][J];        } MAXL=MAXV-MINV;        minl=0;        int flag;            while (1) {mid= (MAXL+MINL) >>1;            flag=0;                    for (l=minv;l+mid<=maxv;l++) if (hungry ()) {flag=1;                Break            } if (flag) Maxl=mid;            if (minl==mid) break;        if (!flag) Minl=mid;    } printf ("%d\n", MAXL); } return 0;}

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Hdu 2236 Untitled II maximum match + two points