2、然后用IDirect3DSurface9的GetDC接口得到dc
3、用dcl图
4、用IDirect3DSurface9的ReleaseDC释放dc
注意Q?/h3>1、buffer的格式必L以下几种之一Q?br />D3DFMT_R5G6B5,D3DFMT_X1R5G5B5,D3DFMT_R8G8B8,D3DFMT_X8R8G8B8
2、D3DPRESENT_PARAMETERS ?Flags需要设|D3DPRESENTFLAG_LOCKABLE_BACKBUFFER
3、GetDC接口在以下情况下会失败：
1Qsurface已经被锁定了
2Qsurface对应的dc没有被释?br />3Qsurface包含在一张texture中，而这个texture中的另一个surface已经被锁定了
4Qsurface存在于default memory poolQƈ且没有设|dynamic usage flag
5Qsurface存在于scratch pool
参考文献：
http://www.xmission.com/~legalize/book/download/04-2D%20Applications.pdf
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//Edmonds-Karp//return the largest flow;flow[] will record every edge's flow//n, the number of nodes in the graph;cap, the capacity //O(VE^2) #define N 100#define inf 0x3f3f3f3fint Edmonds_Karp(int n,int cap[][N],int source,int sink)
{
int flow[N][N]; int pre[N],que[N],d[N]; // d 是增q\长度Qpre 记录前驱Qque是BFS队列 int p,q,t,i,j; if (source==sink) return inf; memset(flow,0,sizeof(flow)); while (true)
{ memset(pre,-1,sizeof(pre)); d[source]=inf; p=q=0, que[q++] = source; while(p < q&&pre[sink]<0) // BFS 找\?/span> { t=que[p++]; for (i=0;i<n;i++) if ( pre[i]<0 && (j=cap[t][i]-flow[t][i]) ) // j取得D余路径?/span> pre[que[q++] = i] = t,d[i] = min(d[t], j);
} if (pre[sink]<0) break; // 找不到增q\Q退?/span> for (i=sink; i!=source; i=pre[i]) { flow[pre[i]][i]+=d[sink]; // 正向量?/span> flow[i][pre[i]]-=d[sink]; // 反向量?/span> } } for (j=i=0; i<n; j+=flow[source][i++]); return j;
}
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#include <stdio.h>#include <memory.h>#define N 1000class treearray{ public: int c[N],n; void clear() { memset(this,0,sizeof(*this)); } int lowbit(int x) { return x&(x^(x-1)); } void change(int i,int d) { for (;i<=n;i+=lowbit(i)) c[i]+=d; } int getsum(int i) { int t; for (t=0;i>0;i-=lowbit(i)) t+=c[i]; return t; }}t;main()//附一个测试程?/span>{ int i,x; t.clear(); scanf("%d",&t.n); for (i=1;i<=t.n;i++) { scanf("%d",&x); t.change(i,x); } for (;scanf("%d",&x),x;) printf("%d\n",t.getsum(x)); return 0;}
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/**//** * TOPSORT(单版) 拓扑排序(Topological Sort) * 输入Q有向图g * 输出Q是否存在拓扑排序，如果存在Q获取拓扑排序序列seq * l构Q图g用邻接矩阵表C?br> * 法Q广度优先搜?BFS) * 复杂度：O(|V|^2) */ #include <iostream>#include <vector>#include <queue>#include <iterator>#include <algorithm>#include <numeric>#include <climits>using namespace std;int n; // n Q顶点个?nbsp;vector<vector<int> > g; // g Q图(graph)(用邻接矩?adjacent matrix)表示) vector<int> seq; // seq Q拓扑序?sequence) bool TopSort(){ vector<int> inc(n, 0); for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) if (g[i][j] < INT_MAX) ++inc[j]; // 计算每个点的入度， queue<int> que; for (int j = 0; j < n; ++j) if (inc[j] == 0) que.push(j); // 如果点的入度ؓ0Q入队?/span> int seqc = 0; seq.resize(n); while (!que.empty()) // 如果队列que非空Q?/span> { int v = que.front(); que.pop(); seq[seqc++] = v; // 点v出队Q放入seq中， for (int w = 0; w < n; ++w) // 遍历所有v指向的顶点wQ?/span> if (g[v][w] < INT_MAX) if (--inc[w] == 0) que.push(w); // 调整w的入度，如果w的入度ؓ0Q入队?nbsp; } return seqc == n; // 如果seq已处理顶Ҏ为nQ存在拓扑排序，否则存在回\?/span>}int main(){ n = 7; g.assign(n, vector<int>(n, INT_MAX)); g[0][1] = 1, g[0][2] = 1, g[0][3] = 1; g[1][3] = 1, g[1][4] = 1; g[2][5] = 1; g[3][2] = 1, g[3][5] = 1, g[3][6] = 1; g[4][3] = 1, g[4][6] = 1; g[6][5] = 1; if (TopSort()) { copy(seq.begin(), seq.end(), ostream_iterator<int>(cout, " ")); cout << endl; } else { cout << "circles exist" << endl; } system("pause"); return 0;}
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// 大整C以一个小整数void big_mul(int d[], int s[], int n){ int plus = 0; for (int i = 1; i < 61; ++i) { d[i] = s[i] * n; d[i] += plus; plus = d[i] / 10; d[i] %= 10; }}// 大整数除以一个小整数void big_div(int d[], int s[], int n){ int left = 0; for (int i = 60; i > 0; --i) { left *= 10; left += s[i]; if (left < n) { d[i] = 0; } else { d[i] = left / n; left %= n; } }}
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/**//*RMQ(Range Minimum/Maximum Query)问题Q?br> RMQ问题是求l定区间中的最值问题。当Ӟ最单的法是O(n)的，但是对于查询ơ数很多Q设|多?00万次Q，O(n)的算法效率不够。可以用U段树将法优化到OQlogn)Q在U段树中保存U段的最|。不q，Sparse_Table法才是最好的Q它可以在O(nlogn)的预处理以后实现O(1)的查询效率。下面把Sparse Table法分成预处理和查询两部分来说明(以求最gؓ??br>预处?预处理用DP的思想Qf(i, j)表示[i, i+2^j - 1]区间中的最|我们可以开辟一个数l专门来保存f(i, j)的倹{?br>例如Qf(0, 0)表示[0,0]之间的最?是num[0], f(0, 2)表示[0, 3]之间的最? f(2, 4)表示[2, 17]之间的最?br>注意, 因ؓf(i, j)可以由f(i, j - 1)和f(i+2^(j-1), j-1)导出, 而递推的初?所有的f(i, 0) = i)都是已知?br>所以我们可以采用自底向上的法递推地给出所有符合条件的f(i, j)的倹{?br>查询:假设要查询从m到nq一D늚最? 那么我们先求Z个最大的k, 使得k满2^k <= (n - m + 1).于是我们可以把[m, n]分成两个(部分重叠?长度?^k的区? [m, m+2^k-1], [n-2^k+1, n];而我们之前已l求Zf(m, k)为[m, m+2^k-1]的最? f(n-2^k+1, k)为[n-2^k+1, n]的最?br>我们只要q回其中更小的那? 是我们惌的答? q个法的时间复杂度是O(1)?例如, rmq(0, 11) = min(f(0, 3), f(4, 3))*/#include<iostream>#include<cmath>using namespace std;#define MAXN 1000000#define mmin(a, b) ((a)<=(b)?(a):(b))#define mmax(a, b) ((a)>=(b)?(a):(b))int num[MAXN];int f1[MAXN][100];int f2[MAXN][100];//试输出所有的f(i, j)void dump(int n){ int i, j; for(i = 0; i < n; i++) { for(j = 0; i + (1<<j) - 1 < n; j++) { printf("f[%d, %d] = %d\t", i, j, f1[i][j]); } printf("\n"); } for(i = 0; i < n; i++) printf("%d ", num[i]); printf("\n"); for(i = 0; i < n; i++) { for(j = 0; i + (1<<j) - 1 < n; j++) { printf("f[%d, %d] = %d\t", i, j, f2[i][j]); } printf("\n"); } for(i = 0; i < n; i++) printf("%d ", num[i]); printf("\n");}//sparse table法void st(int n){ int i, j, k, m; k = (int) (log((double)n) / log(2.0)); for(i = 0; i < n; i++) { f1[i][0] = num[i]; //递推的初?/span> f2[i][0] = num[i]; } for(j = 1; j <= k; j++) { //自底向上递推 for(i = 0; i + (1 << j) - 1 < n; i++) { m = i + (1 << (j - 1)); //求出中间的那个?/span> f1[i][j] = mmax(f1[i][j-1], f1[m][j-1]); f2[i][j] = mmin(f2[i][j-1], f2[m][j-1]); } }}//查询i和j之间的最?注意i是从0开始的void rmq(int i, int j) { int k = (int)(log(double(j-i+1)) / log(2.0)), t1, t2; //用对2d数的Ҏ求出k t1 = mmax(f1[i][k], f1[j - (1<<k) + 1][k]); t2 = mmin(f2[i][k], f2[j - (1<<k) + 1][k]); printf("%d\n",t1 - t2);}int main(){ int i,N,Q,A,B; scanf("%d %d", &N, &Q); for (i = 0; i < N; ++i) { scanf("%d", num+i); } st(N); //初始?br> //dump(N); //试输出所有f(i, j) while(Q--) { scanf("%d %d",&A,&B); rmq(A-1, B-1); } return 0;}
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