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C++ VVI类代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了C++中VVI的典型用法代码示例。如果您正苦于以下问题:C++ VVI类的具体用法?C++ VVI怎么用?C++ VVI使用的例子?那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。



在下文中一共展示了VVI类的20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,您的评价将有助于我们的系统推荐出更棒的C++代码示例。

示例1: latin_square

 latin_square(const VVI& mat, const VI& rp, const VI& cp) : r(mat.size(), 0),
                                                            c(mat.size(), 0),
                                                            m(mat.size(), VI(mat.size())) {
   int n = mat.size();
   for (int i = 0; i < n; i++) {
     for (int j = 0; j < n; j++) {
       r[i] <<= 3;
       r[i] += mat[rp[i]][cp[j]];
       c[i] <<= 3;
       c[i] += mat[rp[j]][cp[i]];
       m[i][j] = mat[rp[i]][cp[j]];
     }
   }
 }
开发者ID:giusevtr,项目名称:problemsolving,代码行数:14,代码来源:Towers.cpp


示例2: mul

VVI mul(VVI a, VVI b) {
	int m = a.size();
	int n = a[0].size();
	int p = b.size();
	VVI c(m, VI(p, 0));
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < p; j++) {
			for (int k = 0; k < n; k++) {
				c[i][j] += a[i][k] * b[k][j];
			}
			if (c[i][j] > 0) c[i][j] = 1;
		}
	}
	return c;
}
开发者ID:colinlee1999,项目名称:leetcode,代码行数:15,代码来源:547.cpp


示例3: display

void display(VVI v){
	for(int i=0;i<v.size();i++){
		for(int j=0;j<v[i].size();j++)
			printf("%d ", v[i][j]);
		printf("\n");
	}
}
开发者ID:jiteshsunhala,项目名称:spoj-solutions,代码行数:7,代码来源:GR.cpp


示例4: d1s1

void d1s1(VVI &inp){
    int vald1s1 = 0;
    int vald1s2 = 0;
    int vald2s1 = 0;
    int vald2s2 = 0;
    int n=inp.size();
    int n1 = n-1;
    int ij ;
    STACK stid1s1,stjd1s1;
    STACK stid1s2,stjd1s2;
    STACK stid2s1,stjd2s1;
    STACK stid2s2,stjd2s2;
    for(int i=0;i<n;i++){
        for(int j=0;j<=i;j++){
            ij = i-j;
            update(inp,stid1s1,stjd1s1,ij,j,vald1s1);
            if(i<(n1)){
               update(inp,stid1s2,stjd1s2,n1-j,n1-ij,vald1s2);
            }
            update(inp,stid2s1,stjd2s1,j,n1-ij,vald2s1);
            if(i<(n1)){
               update(inp,stid2s2,stjd2s2,n1-ij,j,vald2s2);
            }
        }
        emptyStacks(inp,stid1s1,stjd1s1,vald1s1);
        emptyStacks(inp,stid1s2,stjd1s2,vald1s2);
        emptyStacks(inp,stid2s1,stjd2s1,vald2s1);
        emptyStacks(inp,stid2s2,stjd2s2,vald2s2);
    }
}
开发者ID:nitin-nizhawan,项目名称:incubator,代码行数:30,代码来源:MAXCROSSv2.0.cpp


示例5: CalcLambda

/// Calc diagonal matrix Lambda
void CalcLambda(const VVI RM, VI & LM) {
	if (RM.size() == 0) {
		cerr << "Size of relation matrix!" << endl;
		exit(-1);
	}
	LM.clear();
	unsigned int i, j;
	int sum = 0;
	for (i = 0; i < RM.size(); ++i) {
		sum = 0;
		for (j = 0; j < RM[i].size(); ++j) {
			sum += RM[i][j];
		}
		LM.push_back(sum);
	}
}
开发者ID:karthicks123,项目名称:iRoughSet,代码行数:17,代码来源:CRS2.cpp


示例6: GetMinCut

pair<int, VI> GetMinCut(VVI &weights) {
    int N = weights.size();
    VI used(N), cut, best_cut;
    int best_weight = -1;
    
    for (int phase = N-1; phase >= 0; phase--) {
        VI w = weights[0];
        VI added = used;
        int prev, last = 0;
        for (int i = 0; i < phase; i++) {
            prev = last;
            last = -1;
            for (int j = 1; j < N; j++)
                if (!added[j] && (last == -1 || w[j] > w[last])) last = j;
            if (i == phase-1) {
                for (int j = 0; j < N; j++) weights[prev][j] += weights[last][j];
                for (int j = 0; j < N; j++) weights[j][prev] = weights[prev][j];
                used[last] = true;
                cut.push_back(last);
                if (best_weight == -1 || w[last] < best_weight) {
                    best_cut = cut;
                    best_weight = w[last];
                }
            } else {
                for (int j = 0; j < N; j++)
                    w[j] += weights[last][j];
                added[last] = true;
            }
        }
    }
    return make_pair(best_weight, best_cut);
}
开发者ID:ISeaTeL,项目名称:Codebook,代码行数:32,代码来源:MinCut.cpp


示例7: solve_impl

int solve_impl(VVI &adj_list, int node, int k) {
  int N = adj_list.size();
  static VB visited(N);
  static VI fn(N), gn(N);

  // fn[i] contains no of subtrees rooted at i
  // gn[i] contains no of subtrees not rooted at i

  visited[node] = true;
  VI f_values, g_values;
  for (int i = 0; i < adj_list[node].size(); i++) {
    int child = adj_list[node][i];
    if (!visited[child]) {
      solve_impl(adj_list, child, k);
      f_values.push_back(fn[child]);
      g_values.push_back(gn[child]);
    }
  }
  fn[node] = max(subset_sum_count(f_values, k - 1) + 1, 1);
  gn[node] = 0;
  for (int i = 0; i < f_values.size(); i++) {
    gn[node] += f_values[i] + g_values[i];
  }
  cout << "fn and gn for " << node << " are " << fn[node] << " " << gn[node]
       << endl;
  return fn[node] + gn[node];
}
开发者ID:deepakguptagit,项目名称:practice,代码行数:27,代码来源:solution.cpp


示例8: LCSall

set<VT> LCSall(VT& A, VT& B)
{
  VVI dp;
  int n = A.size(), m = B.size();
  dp.resize(n+1);
  for(int i=0; i<=n; i++) dp[i].resize(m+1, 0);
  for(int i=1; i<=n; i++)
    for(int j=1; j<=m; j++)
    {
      if(A[i-1] == B[j-1]) dp[i][j] = dp[i-1][j-1]+1;
      else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
    }
  set<VT> res;
  backtrackall(dp, res, A, B, n, m);
  return res;
}
开发者ID:Jestter379,项目名称:Competitive-Programming,代码行数:16,代码来源:LCS.cpp


示例9: can

bool can(int prefLength,VI& result,bool return_result = false){
	result.clear();

	vector<Hash> hashes;
	int n = 1;
	for(int sn = 0; sn<n;++sn){
		HashComputer<mod1> curHash1;
		HashComputer<mod2> curHash2;

		for(int i = 0; i < prefLength; ++i){
			curHash1.add(a[sn][i]);
			curHash2.add(a[sn][i]);
		}
		hashes.push_back(Hash(make_pair(curHash1.value,curHash2.value),sn,0));

		for(int i = prefLength; i < a[sn].size(); ++i){
			curHash1.shift(a[sn][i-prefLength],a[sn][i],prefLength);
			curHash2.shift(a[sn][i-prefLength],a[sn][i],prefLength);
			hashes.push_back(Hash(make_pair(curHash1.value,curHash2.value),sn,i-prefLength+1));
		}
	}

	sort(hashes.begin(),hashes.end());

	//hashes.resize(unique(hashes.begin(),hashes.end(),EqualString())-hashes.begin());

	
	VVI results;
	for(int i = k-1; i < hashes.size();++i)
		if(hashes[i].equals(hashes[i-k+1])){
			if(return_result == true){
				results.push_back(VI());
				int k = results.size()-1;
				for(int j = i-n+1; j <=i;++j)
					results[k].push_back(hashes[j].pos);
			}
			else
				return true;
		}

	if(return_result == true){
		result = *min_element(results.begin(),results.end());
		return true;
	}

	return false;
}
开发者ID:Yan-Song,项目名称:burdakovd,代码行数:47,代码来源:i.cpp


示例10: print

void print(VVI& v) {
    for(int i = 0; i < v.size(); ++i) {
        for(int j = 0; j < v[i].size() - 1; ++j) {
            cout<<v[i][j]<<" ";
        }
        cout<<endl;
    }
}
开发者ID:xci,项目名称:Tiralabra,代码行数:8,代码来源:slowexact.cpp


示例11: HLD

	HLD(VVI adj) {
		N = adj.size();
		P = PC = VI(N , -1);
		LVL = SZ = groupId = groupPos = VI(N , 0);
		this->adj = adj;
		for(int i = 0; i < N; ++i)
		{
			if(SZ[i] == 0)
			{
				initDfs(i);
				groups.push_back(VI());
				for(int j = i; j != -1; j = PC[j])
					groups.back().push_back(j) ,
					groupId[j] = groups.size() - 1 ,
					groupPos[j] = groups.back().size()-1;
			}
		}
	}
开发者ID:cjtoribio,项目名称:Algorithms,代码行数:18,代码来源:HeavyLightDecomposition.cpp


示例12: OutputMatrix

void OutputMatrix(const VVI matrix) {
	unsigned int i, j;
	for (i = 0; i < matrix.size(); ++i) {
		for (j = 0; j < matrix[i].size(); ++j) {
			cout << matrix[i][j] << " ";
		}
		cout << endl;
	}
}
开发者ID:karthicks123,项目名称:iRoughSet,代码行数:9,代码来源:CRS2.cpp


示例13: main

int main()
{
    int N, M;

    while (cin >> N >> M)
    {
        int a, b;
    
        VVI w;
        VI mr;
        VI mc;
    
        w.resize(N);
    
        for (int i = 0; i < N; ++i)
        {
            w[i].resize(N);
            
            for (int j = 0; j < N; ++j)
            {
                w[i][j] = 0;
            }
        }
    
        for (int i = 0; i < M; ++i)
        {
            cin >> a >> b;
            w[a][b] = 1;
        }
        
        int numMatches = BipartiteMatching(w, mr, mc);
        
        if (numMatches == N)
        {
            cout << "YES\n";
        }
        else
        {
            cout << "NO\n";
        }
    }

    return 0;
}
开发者ID:rhaps0dy,项目名称:UPF-SWERC,代码行数:44,代码来源:12880_bookclub.cpp


示例14: imprime

void imprime(VVI& dist){
	const int V = dist.size();
	for(int i=0;i<V;++i){
		for(int j=0;j<V;++j){
			if(dist[i][j]==INF) cout<<"- ";
			else cout<<dist[i][j]<< " ";
		}
		cout<<endl;
	}
}
开发者ID:wesklei,项目名称:TEG,代码行数:10,代码来源:Trab_03_Gabriel_Wesklei.cpp


示例15: LCS

VT LCS(VT& A, VT& B)
{
  VVI dp;
  int n = A.size(), m = B.size();
  dp.resize(n+1);
  for(int i=0; i<=n; i++) dp[i].resize(m+1, 0);
  
  for(int i=1; i<=n; i++)
    for(int j=1; j<=m; j++)
    {
      if(A[i-1] == B[j-1]) dp[i][j] = dp[i-1][j-1]+1;
      else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
    }
    
  VT res;
  backtrack(dp, res, A, B, n, m);
  reverse(res.begin(), res.end());
  return res;
}
开发者ID:Jestter379,项目名称:Competitive-Programming,代码行数:19,代码来源:LCS.cpp


示例16: transpose

VVI transpose(VVI v){
	VVI result;
	result.resize(v.size());
	for(int i=0;i<result.size();i++)
		result[i].resize(v.size());
	for(int i=0;i<v.size();i++){
		for(int j=0;j<v.size();j++){
			result[j][v.size() - 1 - i] = v[i][j];
		}
	}
	return result;
}
开发者ID:jiteshsunhala,项目名称:spoj-solutions,代码行数:12,代码来源:GR.cpp


示例17: check

bool check(const VVI &g)
{
	int n = g.size();
	allisok = 1;
	for (int i = 0; i < n; ++i)
		colors[i] = 0;
	for (int i = 0; i < n; ++i)
		if (colors[i] == 0)
			dfs(g, i, 1);
	return allisok;
}
开发者ID:dima74,项目名称:Seminars,代码行数:11,代码来源:3v3fast.cpp


示例18: sum_filcol

//SUMA DE MATRICES BAJO EL ENFOQUE FILA-COLUMNA
VVI sum_filcol(VVI A, VVI B)
{

	int size=A.size(); 
	VVI rslt(size,VI(size));
	
	for(int i=0;i<size;i++)	
		for(int j=0;j<size;j++)
			rslt[i][j]=A[i][j]+B[i][j];
	return rslt;
	
}
开发者ID:amh28,项目名称:operaciones-en-memoria-cache,代码行数:13,代码来源:naive.cpp


示例19: GetMaxFlow

 int GetMaxFlow(int s, int t) {
   N = cap.size();
   flow = VVI(N, VI(N));
   reached = VI(N);
   
   int totflow = 0;
   while (int amt = Augment(s, t, INF)) {
     totflow += amt;
     fill(reached.begin(), reached.end(), 0);
   }
   return totflow;
 }
开发者ID:ksalwa,项目名称:Algorithms,代码行数:12,代码来源:GraphCutInference.cpp


示例20: sum_colfil

//SUMA DE MATRICES BAJO EL ENFOQUE COLUMNA-FILA
VVI sum_colfil(VVI A, VVI B)
{

	int size=A.size(); 
	VVI rslt(size,VI(size));

	for(int i=0;i<size;i++)	
		for(int j=0;j<size;j++)
			rslt[j][i]=A[j][i]+B[j][i];
	return rslt;
	
}
开发者ID:amh28,项目名称:operaciones-en-memoria-cache,代码行数:13,代码来源:naive.cpp



注:本文中的VVI类示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


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