5.1 基础题目选讲

11624 - Fire!

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2671

bfs预处理出每个点起火时间，然后再bfs出去到时间。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #define mt(a,b) memset(a,b,sizeof(a))
  5 using namespace std;
  6 const int inf=0x3f3f3f3f;
  7 const int M=1024;
  8 int t,n,m;
  9 char a[M][M];
 10 int d[M][M];
 11 int dx[]={0,0,1,-1};
 12 int dy[]={1,-1,0,0};
 13 struct Q{
 14     int x,y,step;
 15 }now,pre;
 16 queue<Q> q;
 17 bool inside(const int &x,const int &y){
 18     if(x>=0&&x<n&&y>=0&&y<m) return true; return false;
 19 }
 20 void bfs(){
 21     for(int i=0;i<n;i++){
 22         for(int j=0;j<m;j++){
 23             d[i][j]=inf;
 24         }
 25     }
 26     while(!q.empty()) q.pop();
 27     for(int i=0;i<n;i++){
 28         for(int j=0;j<m;j++){
 29             if(a[i][j]=='F'){
 30                 now.x=i;
 31                 now.y=j;
 32                 now.step=0;
 33                 d[i][j]=0;
 34                 q.push(now);
 35             }
 36         }
 37     }
 38     while(!q.empty()){
 39         pre=q.front();
 40         q.pop();
 41         for(int i=0;i<4;i++){
 42             int tx=pre.x+dx[i];
 43             int ty=pre.y+dy[i];
 44             if(!inside(tx,ty)||a[tx][ty]=='#') continue;
 45             now.step=pre.step+1;
 46             if(d[tx][ty]>now.step){
 47                 d[tx][ty]=now.step;
 48                 now.x=tx;
 49                 now.y=ty;
 50                 q.push(now);
 51             }
 52         }
 53     }
 54 }
 55 bool out(const Q &a){
 56     if(a.x==0||a.x==n-1||a.y==0||a.y==m-1) return true; return false;
 57 }
 58 bool vis[M][M];
 59 int solve(){
 60     mt(vis,0);
 61     while(!q.empty()) q.pop();
 62     for(int i=0;i<n;i++){
 63         for(int j=0;j<m;j++){
 64             if(a[i][j]=='J'){
 65                 vis[i][j]=true;
 66                 now.x=i;
 67                 now.y=j;
 68                 now.step=0;
 69                 q.push(now);
 70             }
 71         }
 72     }
 73     while(!q.empty()){
 74         pre=q.front();
 75         q.pop();
 76         if(out(pre)) return pre.step+1;
 77         for(int i=0;i<4;i++){
 78             int tx=pre.x+dx[i];
 79             int ty=pre.y+dy[i];
 80             if(!inside(tx,ty)||a[tx][ty]=='#'||vis[tx][ty]||d[tx][ty]<=pre.step+1) continue;
 81             vis[tx][ty]=true;
 82             now.x=tx;
 83             now.y=ty;
 84             now.step=pre.step+1;
 85             q.push(now);
 86         }
 87     }
 88     return -1;
 89 }
 90 int main(){
 91     while(~scanf("%d",&t)){
 92         while(t--){
 93             scanf("%d%d",&n,&m);
 94             for(int i=0;i<n;i++){
 95                 scanf("%s",a[i]);
 96             }
 97             bfs();
 98             int ans=solve();
 99             if(~ans) printf("%d\n",ans);
100             else puts("IMPOSSIBLE");
101         }
102     }
103     return 0;
104 }

View Code

10047 - The Monocycle

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=988

bfs 每个位置x，y 有朝向d，有颜色c，状态是四维的，有三种走法，左右前。

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<queue>
  4 using namespace std;
  5 const int inf=0x3f3f3f3f;
  6 const int M=32;
  7 int n,m;
  8 char a[M][M];
  9 int dp[M][M][8][4];
 10 struct Q{
 11     int x,y,step,color,dir;
 12 }now,pre;
 13 queue<Q> q;
 14 int dx[]={-1,0,1,0};
 15 int dy[]={0,1,0,-1};
 16 Q turnright(const Q &pre){
 17     Q now;
 18     now.color=pre.color;
 19     now.x=pre.x;
 20     now.y=pre.y;
 21     now.dir=(pre.dir+1)%4;
 22     now.step=pre.step+1;
 23     return now;
 24 }
 25 Q turnleft(const Q &pre){
 26     Q now;
 27     now.color=pre.color;
 28     now.x=pre.x;
 29     now.y=pre.y;
 30     now.dir=pre.dir-1;
 31     if(now.dir==-1) now.dir=3;
 32     now.step=pre.step+1;
 33     return now;
 34 }
 35 Q ahead(const Q &pre){
 36     Q now;
 37     now.color=(pre.color+1)%5;
 38     now.x=pre.x+dx[pre.dir];
 39     now.y=pre.y+dy[pre.dir];
 40     now.dir=pre.dir;
 41     now.step=pre.step+1;
 42     return now;
 43 }
 44 bool inside(const Q &now){
 45     if(now.x>=0&&now.x<n&&now.y>=0&&now.y<m) return true; return false;
 46 }
 47 void go(const Q &now){
 48     if(!inside(now)||a[now.x][now.y]=='#') return ;
 49     if(dp[now.x][now.y][now.color][now.dir]>now.step){
 50         dp[now.x][now.y][now.color][now.dir]=now.step;
 51         q.push(now);
 52     }
 53 }
 54 void bfs(){
 55     for(int i=0;i<n;i++){
 56         for(int j=0;j<m;j++){
 57             for(int k=0;k<5;k++){
 58                 for(int u=0;u<4;u++){
 59                     dp[i][j][k][u]=inf;
 60                 }
 61             }
 62         }
 63     }
 64     for(int i=0;i<n;i++){
 65         for(int j=0;j<m;j++){
 66             if(a[i][j]=='S'){
 67                 now.x=i;
 68                 now.y=j;
 69             }
 70         }
 71     }
 72     dp[now.x][now.y][0][0]=0;
 73     now.color=0;
 74     now.dir=0;
 75     now.step=0;
 76     while(!q.empty()) q.pop();
 77     q.push(now);
 78     while(!q.empty()){
 79         pre=q.front();
 80         q.pop();
 81         now=turnright(pre);
 82         go(now);
 83         now=turnleft(pre);
 84         go(now);
 85         now=ahead(pre);
 86         go(now);
 87     }
 88 }
 89 int main(){
 90     int cas=1;
 91     while(~scanf("%d%d",&n,&m),n|m){
 92         for(int i=0;i<n;i++){
 93             scanf("%s",a[i]);
 94         }
 95         bfs();
 96         int ex,ey;
 97         for(int i=0;i<n;i++){
 98             for(int j=0;j<m;j++){
 99                 if(a[i][j]=='T'){
100                     ex=i;
101                     ey=j;
102                 }
103             }
104         }
105         if(cas>1) puts("");
106         printf("Case #%d\n",cas++);
107         int ans=inf;
108         for(int i=0;i<4;i++){
109             ans=min(ans,dp[ex][ey][0][i]);
110         }
111         if(ans<inf){
112             printf("minimum time = %d sec\n",ans);
113         }
114         else{
115             puts("destination not reachable");
116         }
117     }
118     return 0;
119 }

View Code

10054 - The Necklace

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=995

把颜色当成结点，一个珍珠的两种颜色就连一条边，求无向图欧拉回路，

无向图存在欧拉回路的条件是每个点的出度入度都是偶数

 1 #include<cstdio>
 2 #include<cstring>
 3 #define mt(a,b) memset(a,b,sizeof(a))
 4 const int M=64;
 5 int g[M][M],du[M];
 6 void dfs(int u){
 7     for(int i=1;i<=50;i++){
 8         if(g[u][i]){
 9             g[u][i]--;
10             g[i][u]--;
11             dfs(i);
12             printf("%d %d\n",i,u);
13         }
14     }
15 }
16 int main(){
17     int t,n,u,v;
18     while(~scanf("%d",&t)){
19         for(int cas=1;cas<=t;cas++){
20             scanf("%d",&n);
21             mt(g,0);
22             mt(du,0);
23             while(n--){
24                 scanf("%d%d",&u,&v);
25                 g[u][v]++;
26                 g[v][u]++;
27                 du[u]++;
28                 du[v]++;
29             }
30             bool flag=false;
31             for(int i=1;i<=50;i++){
32                 if(du[i]&1){
33                     flag=true;
34                     break;
35                 }
36             }
37             if(cas>1) puts("");
38             printf("Case #%d\n",cas);
39             if(flag){
40                 puts("some beads may be lost");
41                 continue;
42             }
43             dfs(u);
44         }
45     }
46     return 0;
47 }

View Code

4255 - Guess

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2256&mosmsg=Submission+received+with+ID+1596529

ans[]记录前n项和，+号说明Ax+...Ay>0 ==> Sumy-Sum(x-1)>0 ==> Sumy>Sum(x-1) 负号同理，就能得到前n项和之间的大小关系，可以根据大于的关系建一条有向边，然后边拓扑排序边更新ans。这里采用的是至少要少1,因为答案不唯一。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #define mt(a,b) memset(a,b,sizeof(a))
 5 using namespace std;
 6 const int M=32;
 7 char a[M];
 8 struct G{
 9     struct E{
10         int v,next;
11     }e[M*M];
12     int le,head[M];
13     void init(){
14         le=0;
15         mt(head,-1);
16     }
17     void add(int u,int v){
18         e[le].v=v;
19         e[le].next=head[u];
20         head[u]=le++;
21     }
22 }g;
23 int t,n,rudu[M],ans[M];
24 queue<int> q;
25 void toposort(){
26     mt(ans,0);
27     while(!q.empty()) q.pop();
28     for(int i=0;i<=n;i++){
29         if(!rudu[i]) q.push(i);
30     }
31     while(!q.empty()){
32         int u=q.front();
33         q.pop();
34         for(int i=g.head[u];~i;i=g.e[i].next){
35             int v=g.e[i].v;
36             ans[v]=min(ans[v],ans[u]-1);
37             rudu[v]--;
38             if(!rudu[v]){
39                 q.push(v);
40             }
41         }
42     }
43 }
44 int main(){
45     while(~scanf("%d",&t)){
46         while(t--){
47             scanf("%d%s",&n,a);
48             int p=0;
49             g.init();
50             mt(rudu,0);
51             for(int i=1;i<=n;i++){
52                 for(int j=i;j<=n;j++){
53                     if(a[p]=='-'){
54                         g.add(i-1,j);
55                         rudu[j]++;
56                     }
57                     else if(a[p]=='+'){
58                         g.add(j,i-1);
59                         rudu[i-1]++;
60                     }
61                     p++;
62                 }
63             }
64             toposort();
65             for(int i=1;i<=n;i++){
66                 printf("%d ",ans[i]-ans[i-1]);
67             }
68             puts("");
69         }
70     }
71     return 0;
72 }

View Code

vector存图

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 #include<queue>
 5 #define mt(a,b) memset(a,b,sizeof(a))
 6 using namespace std;
 7 const int M=32;
 8 char a[M];
 9 vector<int> g[M];
10 int t,n,rudu[M],ans[M];
11 queue<int> q;
12 void toposort(){
13     mt(ans,0);
14     while(!q.empty()) q.pop();
15     for(int i=0;i<=n;i++){
16         if(!rudu[i]) q.push(i);
17     }
18     while(!q.empty()){
19         int u=q.front();
20         q.pop();
21         int len=g[u].size();
22         for(int i=0;i<len;i++){
23             int v=g[u][i];
24             ans[v]=min(ans[v],ans[u]-1);
25             rudu[v]--;
26             if(!rudu[v]){
27                 q.push(v);
28             }
29         }
30     }
31 }
32 int main(){
33     while(~scanf("%d",&t)){
34         while(t--){
35             scanf("%d%s",&n,a);
36             int p=0;
37             for(int i=0;i<=n;i++) g[i].clear();
38             mt(rudu,0);
39             for(int i=1;i<=n;i++){
40                 for(int j=i;j<=n;j++){
41                     if(a[p]=='-'){
42                         g[i-1].push_back(j);
43                         rudu[j]++;
44                     }
45                     else if(a[p]=='+'){
46                         g[j].push_back(i-1);
47                         rudu[i-1]++;
48                     }
49                     p++;
50                 }
51             }
52             toposort();
53             for(int i=1;i<=n;i++){
54                 printf("%d ",ans[i]-ans[i-1]);
55             }
56             puts("");
57         }
58     }
59     return 0;
60 }

View Code

end

posted on 2014-11-03 14:44 gaolzzxin 阅读(149) 评论(0) 编辑收藏举报