1.6二分法
1 题目：有n张纸片，随机取4张（可放回），如4张面值加起来可等于m，则输出yes，否则no。纸片的面值为k[1],k[2]……
2
3 思路：使用4次循环寻找会导致超时，所以假设4张分别为a,b,c,d，
4
5 先计算二次循环a+b所有可能放入数组kk，然后将kk排序，
6
7 最后使用二次循环（m-c-d）与kk用二分法比较，，最后确定是否有这个值。
8
9 #include <stdio.h>
10 #include <algorithm>
11 using namespace std;
12 bool ss(int x);
13 void solve();
14 int n,m,r,k[500];
15 int kk[500];
16 bool f;
17 int main()
18 {
19 while(scanf("%d %d",&n,&m)!=EOF)
20 {
21 for(int i=0;i<n;i++)
22 scanf("%d",&k[i]);
23 solve();
24 if(f) printf("Yes\n");
25 else printf("NO\n");
26 }
27 return 0;
28 }
29
30 void solve()
31 {
32 for(int a=0;a<n;a++)
33 for(int b=0;b<n;b++)
34 kk[a*n+b]=k[a]+k[b];
35 sort(kk,kk+n*n);
36 f=false;
37 for(int c=0;c<n;c++)
38 for(int d=0;d<n;d++)
39 if(ss(m-k[c]-k[d]))
40 {
41 f=true;
42 }
43 }
44
45 bool ss(int x)
46 {
47 int l=0;r=n*n;
48 while(r-l>=1)
49 {
50 int i=(r+l)/2;
51 if(kk[i]>x)l=i+1;
52 else if (kk[i]==x) return true;
53 else r=i;
54 }
55 }
56
57
58 Tip:此题也可以使用检索a然后排序，然后（m-b-c-d）与a用二分法比较，套路差不多，但是第一种方法的思想明显比较高大上。
59
60 对了，局部变量可以与全局变量重名，但是局部变量会屏蔽全局变量哦！！
1.6
2.1穷竭搜索
DFS
1 /*
2 input：
3 7 7
4 ..#.#..
5 ..#.#..
6 ###.###
7 [email protected]
8 ###.###
9 ..#.#..
10 ..#.#.. 表示有13个可以走到的地方（包括原来那一格）。*/
11
12
13 #include <stdio.h>
14 char ch[24][24];
15 void solve(int H,int W);
16 void find1(int H,int W);
17 void dfs(int x,int y,int H,int W);
18 int main()
19 {
20 int H,W,i;
21 while(scanf("%d%d",&W,&H)!=EOF&&H!=0&&W!=0)
22 {
23 getchar();
24 for(i=0;i<H;i++)
25 gets(ch[i]);
26 find1(H,W);
27 solve(H,W);
28 }
29 return 0;
30 }
31
32 void solve(int H,int W)
33 {
34 int a=0;
35 for(int i=0;i<H;i++)
36 for(int k=0;k<W;k++)
37 if(ch[i][k]==‘@‘)
38 a++;
39 printf("%d\n",a);
40 }
41
42 void find1(int H,int W)
43 {
44 for(int i=0;i<H;i++)
45 for(int k=0;k<W;k++)
46 if(ch[i][k]==‘@‘)
47 {
48 dfs(i,k,H,W);
49 return;
50 }
51 }
52
53 void dfs(int x,int y,int H,int W)
54 {
55 int x1,y1,nx,ny;
56 ch[x][y]=‘@‘;
57 for(x1=-1;x1<=1;x1++)
58 {
59 nx=x+x1;
60 if(nx>=0&&nx<H&&ch[nx][y]!=‘#‘&&ch[nx][y]!=‘@‘)
61 dfs(nx,y,H,W);
62 }
63 for(y1=-1;y1<=1;y1++)
64 {
65 ny=y+y1;
66 if(ny>=0&&ny<W&&ch[x][ny]!=‘#‘&&ch[x][ny]!=‘@‘)
67 dfs(x,ny,H,W);
68 }
69 return;
70 }
71
72 POJ2386
1 #include <stdio.h>
2 char ch[24][24];
3 void solve(int H,int W);
4 void find1(int H,int W);
5 void dfs(int x,int y,int H,int W);
6 int main()
7 {
8 int H,W,i;
9 while(scanf("%d%d",&W,&H)!=EOF&&H!=0&&W!=0)
10 {
11 getchar();
12 for(i=0;i<H;i++)
13 gets(ch[i]);
14 find1(H,W);
15 solve(H,W);
16 }
17 return 0;
18 }
19
20 void solve(int H,int W)//计数
21 {
22 int a=0;
23 for(int i=0;i<H;i++)
24 for(int k=0;k<W;k++)
25 if(ch[i][k]==‘@‘)
26 a++;
27 printf("%d\n",a);
28 }
29
30 void find1(int H,int W)//找到点
31 {
32 for(int i=0;i<H;i++)
33 for(int k=0;k<W;k++)
34 if(ch[i][k]==‘@‘)
35 {
36 dfs(i,k,H,W);
37 return;
38 }
39 }
40
41 void dfs(int x,int y,int H,int W)
42 {
43 int x1,y1,nx,ny;
44 ch[x][y]=‘@‘;
45 for(x1=-1;x1<=1;x1++)
46 {
47 nx=x+x1;
48 if(nx>=0&&nx<H&&ch[nx][y]!=‘#‘&&ch[nx][y]!=‘@‘)
49 dfs(nx,y,H,W);
50 }
51 for(y1=-1;y1<=1;y1++)
52 {
53 ny=y+y1;
54 if(ny>=0&&ny<W&&ch[x][ny]!=‘#‘&&ch[x][ny]!=‘@‘)
55 dfs(x,ny,H,W);
56 }
57 return;
58 }
59
60 POJ1979
1 #include <stdio.h>
2 #include <iostream>
3 using namespace std;
4 char ch[102][102];
5 int xx[4]={0,1,0,-1};
6 int yy[4]={1,0,-1,0};
7 int H,W,cou;
8 void DFS(int x, int y){
9 int a=0;
10 char b=ch[x][y];
11 ch[x][y]=‘.‘;
12 for(int i = 0; i < 4; i++)
13 {
14 int xn = x + xx[i], yn = y + yy[i];
15 if(xn >= 0 && xn < W && yn >= 0 && yn < H && ch[xn][yn] == b)
16 {
17 DFS(xn, yn);
18 }
19 }
20 return ;
21 }
22
23 int main()
24 {
25 while(cin >> W >> H && (H && W))
26 {
27 cou = 0;
28 for(int i=0; i < W; i++)
29 for(int k=0; k < H; k++)
30 cin >> ch[i][k];
31 for(int i = 0; i < W; i++)
32 for(int k = 0; k < H; k++)
33 if(ch[i][k] != ‘.‘)
34 {
35 DFS(i, k);
36 cou++;
37 }
38 cout << cou << endl;
39 }
40 return 0;
41 }
42
43 AOJ0033
1 #include <iostream>
2 #include <string.h>
3 #include <algorithm>
4 using namespace std;
5 int dx[4]={ 0, 1, -1, 0};
6 int dy[4]={ 1, 0, 0, -1};
7 int map1[22][22];
8 int xn,yn,x1,y1,H,W,y2,x2,a;
9 void dfs(int x1, int y1, int step){
10 if(step > 10) return ;
11 for(int i = 0; i < 4; i++){//四个方向
12 int xn = x1 + dx[i], yn = y1 + dy[i];
13 if(xn == x2 && yn == y2){
14 a=min(a, step + 1);
15 continue;
16 }
17 if(map1[xn][yn] == 1)continue;
18 while(xn >= 0 && xn < W && yn >= 0 && yn < H){
19 xn += dx[i]; yn += dy[i];//继续往下面走
20 if(xn >= 0 && xn < W && yn >= 0 && yn < H)
21 {
22 if(map1[xn][yn] == 0) continue;//直到撞墙或终点
23 if(xn == x2 && yn == y2){//到达终点
24 a= min(a, step + 1);
25 return ;
26 }
27 map1[xn][yn] = 0;//停下来了，继续dfs，注意dfs前后面的墙。
28 dfs(xn - dx[i], yn - dy[i], step + 1);
29 map1[xn][yn] = 1;
30 break;
31 }
32 }
33 }
34 return ;
35 }
36
37 int main()
38 {
39 while(cin >> H >> W && H != 0 && W != 0){
40 for(int i = 0; i < W; i++)
41 for(int k = 0; k < H; k++){
42 cin >> map1[i][k];
43 if(map1[i][k] == 2){
44 x1 = i; y1 = k;
45 map1[i][k] = 0;
46 }
47 if(map1[i][k] == 3){
48 x2 = i;
49 y2 = k;
50 map1[i][k] = 1;
51 }
52 }
53 a = 99;
54 dfs(x1, y1, 0);
55 if(a == 99 || a>10) cout << "-1" << endl;
56 else cout << a <<endl;
57 }
58 return 0;
59 }
BFS
1 #include <stdio.h>
2 #include <queue>
3 #include <string.h>
4 using namespace std;
5 int map1[305][305],step[305][305];
6 typedef pair<int, int> P;
7 struct stu{
8 int x; int y; int t;
9 };
10 struct stu ch[50002];
11
12 int dx[5]={0, 1, 0, -1, 0};
13 int dy[5]={0, 0, 1, 0, -1};
14
15 void zha(int M) {
16 memset(map1, -1, sizeof(map1));
17 for(int i = 0; i < M; i++)
18 for(int k = 0; k < 5; k++){//five points are burst.
19 int nx = ch[i].x + dx[k];
20 int ny = ch[i].y + dy[k];
21 if(nx >= 0 && nx < 303 && ny >= 0 && ny < 303 && (ch[i].t < map1[ny][nx] || map1[ny][nx] == -1))
22 map1[ny][nx] = ch[i].t;//give everypoint the minimum burst time.
23 }
24 }
25
26 int bfs() {
27 queue <P> que;
28 que.push( P (0, 0));
29 memset(step, -1, sizeof(step));
30 step[0][0]=0;
31 while(que.size()) {
32 P p= que.front();
33 que.pop();
34 if(map1[p.first][p.second] == -1){//find map1‘s -1 and print it
35 return step[p.first][p.second];
36 }
37 if(step[p.first][p.second] >= map1[p.first][p.second])continue;//the point had benn burst,so jump it
38 for(int i = 1; i < 5; i++) {//this is the point of four diretions that does not been destroy.
39 int nx = p.first + dx[i],ny = p.second + dy[i];
40 if(nx>=0 && ny >=0 &&step[nx][ny] == -1)//if it can arrive and does not exceed the map
41 {
42 que.push(P(nx, ny));//push it to the queue and go
43 step[nx][ny]=step[p.first][p.second]+1;//of course,step must add 1 when go.
44 }
45 }
46 }
47 return -1;
48 }
49
50 int main()
51 {
52 int M,a;
53 while(~scanf("%d",&M))
54 {
55 for(int i=0; i<M; i++)
56 scanf("%d%d%d",&ch[i].x,&ch[i].y,&ch[i].t);
57 zha(M);
58 a=bfs();
59 printf("%d\n",a);
60 }
61 return 0;
62 }
1 #include <iostream>
2 #include <queue>
3 #include <string.h>
4 using namespace std;
5 typedef pair<int, int> P;
6 int sx,sy,H,W,N;
7 char map1[1002][1002];
8 int step[1002][1002];
9 int dx[4] = {1,0,-1,0};
10 int dy[4] = {0,1,0,-1};
11 void find1(int g){
12 for(int i = 0; i < W; i++)
13 for(int k = 0; k < H; k++)
14 if(map1[i][k] == ‘S‘ || map1[i][k] == (g+48)){
15 if(map1[i][k] == ‘S‘) map1[i][k] =‘.‘;
16 sx = i; sy = k;
17 return ;
18 }
19 }
20
21 int bfs(int n)
22 {
23 memset(step,-1,sizeof(step));
24 queue <P> que;
25 que.push(P (sx,sy));
26 step[sx][sy] = 0;
27 while(que.size()){
28 int x = que.front().first, y = que.front().second;
29 que.pop();
30 for(int i = 0; i < 4; i++){
31 int xn = x + dx[i], yn = y + dy[i];
32 if(xn >= 0 && xn < W && yn >= 0 && yn < H && map1[xn][yn] != ‘X‘ && step[xn][yn] == -1)
33 {
34 step[xn][yn] = step[x][y] + 1;
35 if(map1[xn][yn] == (n+48))
36 {
37 return step[xn][yn];
38 }
39 que.push(P(xn, yn));
40 }
41 }
42 }
43 return 0;
44 }
45
46 int main()
47 {
48 cin >> W >> H >> N;
49 for(int i = 0; i < W; i++)
50 for(int k = 0; k < H; k++)
51 cin >> map1[i][k];
52 int step = 0;
53 find1(0);
54 step += bfs(1);
55 for(int i = 1; i < N; i++)
56 {
57
58 find1(i);
59 step += bfs(i+1);
60 }
61 cout << step << endl;
62 return 0;
63 }
1 #include <iostream>
2 #include <queue>
3 #include <string.h>
4 #include <map>
5 #include <algorithm>
6 using namespace std;
7
8 int d[4]={ 1, -1, 4, -4};
9 map<string, int> dp;
10
11 void bfs(){
12 queue<string> que;
13 que.push("01234567");
14 dp["01234567"] = 0;
15 while(que.size()){
16 string now =que.front();
17 que.pop();
18 int p=0;
19 for(int j = 0; j < 8; j++)
20 if(now[j] == ‘0‘){
21 p=j;
22 break;
23 }
24 for(int i = 0; i < 4; i++){
25 int pn= p + d[i];
26 if(pn >= 0 && pn < 8 && !(p == 3 && i == 0)
27 && !(p ==4 && i == 1)){
28 string next = now;
29 swap(next[p],next[pn]);
30 if(dp.find(next) == dp.end()){
31 dp[next] = dp[now] + 1;
32 que.push(next);
33 }
34 }
35 }
36 }
37 return ;
38 }
39
40 int main()
41 {
42 string line;
43 bfs();
44 while(getline(cin,line))
45 {
46 line.erase(remove(line.begin(), line.end(), ‘ ‘), line.end());
47 cout << dp[line] << endl;
48 }
49 return 0;
50 }
Search
1 #include <iostream>
2 #include <algorithm>
3 #include <stdio.h>
4 #include <string.h>
5 using namespace std;
6 int N,ch[12],sh[12],g,i,cou;
7 bool judge[12];
8
9 void solve(int a){
10 int leng=0,b=0;
11 for(int k = 0;k < i; k++)
12 if(!judge[k]){//如果没被访问过，就加入数组sh和b
13 sh[leng++] = ch[k];
14 b=b * 10 + ch[k];
15 }
16 if(sh[0] != 0 || leng == 1)
17 cou = min(cou,abs(a-b));
18 while(next_permutation(sh,sh+leng)){//（除了原数组）全排列的所有情况
19 b=0;
20 for(int k = 0; k < leng; k++)
21 b = b * 10 + sh[k];
22 if(sh[0] != 0 || leng == 1)
23 {
24 cou = min(cou,abs(a-b));
25 }
26 }
27 return ;
28 }
29
30 void dfs(int x, int y){
31 if(x == i/2){
32 solve(y);
33 return;
34 }
35 for(int k=0; k < i; k++){
36 if(!judge[k]){
37 if(ch[k] == 0 && x == 0 && i > 3)
38 continue;
39 judge[k]=true;//被访问过
40 dfs(x + 1, y * 10 + ch[k]);
41 judge[k]=false;
42 }
43 }
44 return ;
45 }
46
47
48 int main()
49 {
50 cin >> N;
51 getchar();
52 while(N--){
53 for( i = 0;;){
54 cin>> ch[i++];
55 if(g=getchar() == ‘\n‘)
56 break;
57 }
58 cou=1000000;
59 memset(judge,false,sizeof(judge));
60 dfs(0,0);
61 cout << cou <<endl;
62 }
63 return 0;
64 }
1 #include <iostream>
2 #include <algorithm>
3 using namespace std;
4 int n,sum;
5 int tri[10][10];
6
7 int main()
8 {
9 while(cin >> n >> sum)
10 {
11 int lie[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
12 do{
13 for(int k = 0; k < n; k++)
14 tri[0][k]=lie[k];//first line
15 for(int i = 1; i < n; i++)
16 for(int g = 0 ; g < n - i; g++)//next line
17 tri[i][g] = tri[i-1][g] + tri[i-1][g+1];
18 if(tri[n-1][0] == sum )
19 break;
20 } while(next_permutation(lie, lie+n));
21
22 for(int j = 0; j < n ; j++){
23 if(j != 0)
24 cout << " ";
25 cout << lie[j];
26 }
27 cout << endl;
28 }
29 return 0;
30 }
1 #include <iostream>
2 using namespace std;
3 int map1[5][5],cou[10000000];
4 int a,n,h;
5 int x1[4] = {0, 1, -1, 0};
6 int y1[4] = {1, 0, 0, -1};
7
8 void dfs(int x, int y, int c, int all){
9 if(c == 6)
10 {
11 cou[n++] = all;
12 for(h = 0; h < n-1; h++)
13 if(all == cou[h])
14 break;
15 if(h == n-1)
16 a++;
17 return ;
18 }
19 for(int i = 0; i < 4; i++)
20 {
21 int xn = x + x1[i], yn = y + y1[i];
22 if(xn >= 0 && xn <= 4 && yn >= 0 && yn <= 4)
23 {
24 dfs(xn, yn, c + 1, all * 10 + map1[xn][yn]);
25 }
26 }
27 return ;
28 }
29
30 int main()
31 {
32 a = 0;
33 n = 0;
34 for(int i = 0; i < 5; i++)
35 for(int k = 0; k < 5; k++)
36 cin >> map1[i][k];
37 for(int i = 0; i < 5; i++)
38 for(int k = 0; k < 5; k++)
39 dfs(i, k, 1, map1[i][k]);
40 cout << a << endl;
41 return 0;
42 }
1 #include <iostream>
2 #include <bitset>
3 using namespace std;
4
5 bitset<10000> map1[10];
6 int r,c,cou;
7
8 void dfs(int all){
9 if(all == r)
10 {
11 int num, upcou=0;
12 for(int i = 0; i < c; i++)
13 {
14 num=0;
15 for(int k = 0; k < r; k++)//竖向翻转取大的
16 if(map1[k][i])
17 num++;
18 upcou += max(num, r-num);
19 }
20 cou = max(cou, upcou);
21 return;
22 }
23 dfs(all+1);//所有情况
24 map1[all].flip();
25 dfs(all+1);
26 return ;
27 }
28 int main()
29 {
30 while(cin >> r >> c)//r行c列
31 {
32 if(r == 0 && c == 0)
33 break;
34 for(int i = 0,g; i < r; i++)
35 for(int k = 0; k < c; k++)
36 {
37 cin >> g;
38 map1[i][k]=g;
39 }
40 cou = 0;
41 dfs(0);
42 cout << cou <<endl;
43 }
44 return 0;
45 }
2.2 贪心
区间
1 #include <iostream>
2 #include <algorithm>
3 using namespace std;
4
5 const int maxn = 25000;
6 pair<int, int> line[maxn];
7 int g,h,N,T,j,maxitime;
8
9 int main()
10 {
11 int i, t, cou = 0;
12 cin >> N >> T;
13 for(i = 0; i < N; i++)
14 cin >> line[i].first >> line[i].second;
15 sort(line, line + N);//排序，按照开始时间>>结束时间
16
17 if(line[0].first != 1) printf("-1\n");//不是从1开始则失败
18 else
19 {
20 t=line[0].second;
21 i = 1;
22 while(line[i].first == 1) t = line[i++].second;//取得第一段结束时间最晚的
23 cou = 1;
24
25 while(t < T)//后面的段
26 {
27 if(i >= N)break;
28 j = i;
29 maxitime=line[i].second;
30 i++;
31 while( i < N && line[i].first <= t+1)//往后取，直到取尽或者不连续
32 {
33 if(line[i].second > maxitime)
34 {
35 maxitime=line[i].second;//取得每一段结束时间最晚的
36 j=i;//j是有效的段
37 }
38 i++;//往后取
39 }
40 if(maxitime <= t || line[j].first > t+1) break;//不能取完或是非正常区域内的段
41 else
42 {
43 cou++;
44 t = line[j].second;//更新结束时间
45 }
46 }
47 if(t<T) printf("-1\n");
48 else printf("%d\n",cou);
49 }
50 return 0;
51 }
1 #include <iostream>
2 #include <math.h>
3 #include <algorithm>
4 using namespace std;
5
6 bool flag;
7 int ans,T = 0,head,tail,n,d,a,b;
8
9 struct point{
10 double x,y;
11 point(double a = 0, double b = 0){ x = a, y = b; }
12 bool operator < (const point c)const {return x < c.x;}
13 }queue1[1005];
14
15 void push(point v){queue1[tail++]=v;}
16 void pop(){head++;}
17
18 bool judge(point &t,point v){
19 if(!( t.x>v.y || t.y< v.x))
20 {
21 t.x = min(t.x,v.x);
22 t.y = min(t.y,v.y);
23 return true;
24 }
25 return false;
26 }
27
28 int main()
29 {
30 while(cin >> n >> d)
31 {
32 if(!n && !d) break;
33 ans = flag = head = tail = 0;
34 for(int i = 0; i < n; i++)
35 {
36 cin >> a >> b;
37 if(b > d) flag = 1;
38 else
39 {
40 push(point(a-sqrt((double)d*d - b*b),a+sqrt((double)d*d - b*b)));
41 }
42 }
43 if(flag) cout << "Case " << ++T << ": " << "-1" <<endl;
44 else
45 {
46 sort(queue1, queue1 + n);
47 while(head != tail)
48 {
49 ans++;
50 point t = queue1[head];
51 while(head != tail)
52 {
53 point v = queue1[head];
54 if(judge(t,v)) pop();
55 else break;
56 }
57 }
58 cout << "Case " << ++T << ": " << ans <<endl;
59 }
60 }
61 return 0;
62 }
1 #include <iostream>
2 #include <queue>
3 #include <algorithm>
4 using namespace std;
5
6 int N,house[50005];
7
8 struct stu{
9 int a,b,site;
10
11 bool operator < (const stu &a1) const {
12 if(b == a1.b) return a > a1.a;
13 else return b > a1.b;
14 }
15 } line[50005];
16
17 priority_queue<stu> que;
18
19 bool cmp(stu a1, stu b1){
20 if(a1.a > b1.a) return false;
21 if(a1.a < b1.a) return true;
22 else return a1.b < b1.b;
23 }
24
25
26 int main()
27 {
28 while(cin >> N)
29 {
30 for(int i = 0; i < N; i++)
31 {
32 cin >> line[i].a >> line[i].b;
33 line[i].site = i;
34 }
35 sort(line, line + N, cmp);
36
37 int cou=1;
38 house[line[0].site] = 1;
39 que.push(line[0]);
40 for(int i = 1; i < N; i++)
41 {
42 if(!que.empty() && que.top().b < line[i].a)
43 {
44 house[line[i].site] = house[que.top().site];
45 que.pop();
46 }
47 else
48 {
49 cou++;
50 house[line[i].site] = cou;
51 }
52 que.push(line[i]);
53 }
54
55 cout << cou <<endl;
56 for(int i = 0; i < N; i++)
57 cout << house[i] <<endl;
58 while(!que.empty())
59 que.pop();
60 }
61 return 0;
62 }
2.5 图
1 #include <iostream>
2 using namespace std;
3
4 int n,w,m,d[520],num,s,e,t;
5 struct edge{
6 int from, to, cost;
7 }eg[5203];
8
9 bool shortpath(){
10 bool flag;
11 d[1] = 0;
12 for(int i = 1; i < n; i++)
13 {
14 flag = false;
15 for(int j = 0; j < num; j++)
16 if(d[eg[j].from] + eg[j].cost < d[eg[j].to])
17 {
18 d[eg[j].to] = d[eg[j].from] + eg[j].cost;
19 flag = true;
20 }
21 if(!flag)
22 return false;
23 }
24 for(int i = 0; i < num; i++)
25 if(d[eg[i].from] + eg[i].cost < d[eg[i].to])
26 return true;
27
28 }
29
30 int main()
31 {
32 int T;
33 cin >> T;
34 while(T--)
35 {
36 num = 0;
37 cin >> n >> m >> w;
38 for(int i = 0; i < m; i++)
39 {
40 cin >> s >> e >> t;
41 eg[num].from = s;
42 eg[num].to = e;
43 eg[num++].cost = t;
44 eg[num].from = e;
45 eg[num].to = s;
46 eg[num++].cost = t;
47 }
48 for(int i = 0; i < w; i++)
49 {
50 cin >> s >> e >> t;
51 eg[num].from = s;
52 eg[num].to = e;
53 eg[num++].cost = -t;
54 }
55 for(int i = 1; i <= n; i++)
56 d[i]=0x3f3f3f3f;
57 if(shortpath())
58 cout << "YES" <<endl;
59 else cout << "NO" << endl;
60 }
61 return 0;
62 }
时间： 2024-10-06 13:55:33