本文整理汇总了C++中Alignment::to_wstring方法的典型用法代码示例。如果您正苦于以下问题：C++ Alignment::to_wstring方法的具体用法？C++ Alignment::to_wstring怎么用？C++ Alignment::to_wstring使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Alignment的用法示例。
在下文中一共展示了Alignment::to_wstring方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1:
bool
Alignment::unionn(Alignment& al2) {
if (!are_the_same_alignment(al2)) {
wcerr<<L"Error when uniting the following two alignments:\n";
wcerr<<to_wstring()<<L"\n";
wcerr<<al2.to_wstring()<<L"\n";
return false;
}
for(unsigned i=0; i<source.size(); i++) {
for(unsigned j=0; j<target.size(); j++) {
if (al2.alignment[i][j]) {
alignment[i][j]=true;
}
}
}
return true;
}示例2: intersection
bool
Alignment::refined_intersection(Alignment& al2) {
if (!are_the_same_alignment(al2)) {
wcerr<<L"Error when performing the refined intersection of the following two alignments:\n";
wcerr<<to_wstring()<<L"\n";
wcerr<<al2.to_wstring()<<L"\n";
return false;
}
//We save the alignment of (*this) before intersecting
map<int, map<int, bool> > al1;
for(unsigned i=0; i<source.size(); i++) {
for(unsigned j=0; j<target.size(); j++) {
al1[i][j]=alignment[i][j];
}
}
//First, the intersection
intersection(al2);
//cerr<<"(0) "<<to_string()<<"\n";
bool alignments_changed;
//int nit=0;
do {
alignments_changed=false;
//nit++;
for (unsigned i=0; i<source.size(); i++) {
for(unsigned j=0; j<target.size(); j++) {
if (!alignment[i][j]) {
if ((al1[i][j]) || (al2.alignment[i][j])) {
//cerr<<" considering ("<<i<<","<<j<<")\n";
bool add_alignment=true;
for(unsigned k=0; k<target.size(); k++) {
if (alignment[i][k])
add_alignment=false;
}
for(unsigned k=0; k<source.size(); k++) {
if (alignment[k][j])
add_alignment=false;
}
if (!add_alignment) {
//cerr<<" ("<<i<<","<<"*) or (*,"<<j<<") are already in A\n";
//The aligment can still be added if it has an horizontal
//neighbor or a vertical neighbor already in 'alignment'
if ( ((alignment[i-1][j])||(alignment[i+1][j])) ||
((alignment[i][j-1])||(alignment[i][j+1])) ) {
//cerr<<" ("<<i<<","<<j<<") has an horizontal or a vertical neighbor in A\n";
alignment[i][j]=true;
//Now we that test there is no aligments in 'alignment' with
//both horizontal and vertical neighbors
bool neighbors=false;
for(unsigned ii=0; (ii<source.size()) && (!neighbors); ii++) {
for(unsigned jj=0; (jj<target.size()) && (!neighbors); jj++) {
if (alignment[ii][jj]) {
if ( ((alignment[ii-1][jj])||(alignment[ii+1][jj])) &&
((alignment[ii][jj-1])||(alignment[ii][jj+1])) ) {
//cerr<<" ("<<i<<","<<j<<") has both horizontal and vertical neighbors\n";
neighbors=true;
}
}
}
}
alignment[i][j]=false;
if(!neighbors)
add_alignment=true;
}
}
if (add_alignment) {
//cerr<<" adding ("<<i<<","<<j<<")\n";
alignment[i][j]=true;
alignments_changed=true;
}
}
}
}
}
//cerr<<"("<<nit<<") "<<to_string()<<"\n";
} while(alignments_changed);
return true;
}