HDU5093 Battle Ships (two-part picture)

Test instructions: give you a m*n (1<=m,n<=50) figure, where ' # ' represents the iceberg, ' * ' represents the ocean, ' o ' represents the ice floe. Then let you place as many boats as possible, but to satisfy the rules:

1. A ship cannot be placed on an iceberg;
2. The ship cannot be placed on the ice.
3. The two ships cannot be in the same column or row unless there is an iceberg between them.

Analysis: The two best match of red fruits .... The diagram is also very easy to construct ... The Ocean lattice connected block (at least one lattice) separated by an iceberg is used as the X-point, and each column of the same row is separated by an iceberg as a Y-point, and the X-point is connected to the Y-point with an edge, and only if the two connected blocks share a single ocean lattice. Then run the two-point maximum match ... 2,500 points, duly completed.

As if it is the official work, but can not, the next array to Nextt to be able to cross C + +, g++.

#include <stdlib.h> #include <cmath> #include <cstring> #include <iostream> #include <map > #include <set> #include <algorithm> #include <queue> #include <vector> #include <cstdio >using namespace std; #define N 205#define M 50005int ev[m],nextt[m];int head[m];int cnt;bool vis[m];int Pre[m];char S[N    ][n];int x[n][n],y[n][n];int pn[2];void addedge (int u,int v) {ev[cnt]=v;    Nextt[cnt]=head[u];    head[u]=cnt++; return;}        int find (int x) {for (int i=head[x];~i;i=nextt[i]) {int v=ev[i];            if (!vis[v]) {vis[v]=1; if (pre[v]==-1| |                Find (Pre[v])) {pre[v]=x;            return 1; }}} return 0;}    int main () {int t;    scanf ("%d", &t);        while (t--) {int n,m;        memset (head,-1,sizeof (head));        cnt=0;        memset (pre,-1,sizeof (pre));        memset (x,0,sizeof (x));        memset (y,0,sizeof (y));        scanf ("%d%d", &n,&m); for (int i=0;i<n;i++) scanf ("%s", S[i]);        int num=0;            for (int i=0;i<n;i++) {bool flag=0;                    for (int j=0;j<m;j++) {if (s[i][j]== ' * ') {if (flag==0) num++;                    X[i][j]=num;                flag=1;            } else if (s[i][j]== ' # ') flag=0;        }} Pn[0]=num;        num=0;            for (int i=0;i<m;i++) {bool flag=0;                    for (int j=0;j<n;j++) {if (s[j][i]== ' * ') {if (flag==0) num++;                    Y[j][i]=num;                flag=1;            } else if (s[j][i]== ' # ') flag=0;        }} Pn[1]=num;            for (int i=0;i<n;i++) for (int j=0;j<m;j++) {int u = x[i][j],v=y[i][j];        if (u&&v) Addedge (U,V);        } int ans=0; for (int i=1;i<=pn[0];++i) {memset (vis,0,sizeof (VIS));        Ans+=find (i);    } printf ("%d\n", ans); } return 0;}

HDU5093 Battle Ships (two-part picture)