Uva 429 Word Transformation ( 最短路径 )
2013-03-03 14:22
211 查看
这道题一直是TLE,超时,开始以为是算法的问题,后来才知道是输入的问题,太坑了
这道题可以用BFS来解, floyd也没有问题
但是一定要注意,输入单词对的时候,结束符分两种情况,一种是 几组连续输入中间的是空行为结束,最后一组数据是以EOF结束
注:注释掉的 有一部分是floyd算,但是一下正常的是用BFS解题
代码:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <map>
#include <queue>
using namespace std;
const int N = 300;
const int INF = 1000000;
int T, id;
int g
, d
;
map <string, int> mp;
string s1, s2, s
;
int bfs( int st, int en ) {
queue< int > q;
q.push(st);
for ( int i = 0; i <= id; ++i ) d[i] = INF;
d[st] = 0;
while ( !q.empty() ) {
int u = q.front(); q.pop();
if ( u == en ) break;
for ( int i = 0; i < id; ++i ) if ( g[u][i] && d[i] > d[u] + 1 ) {
q.push(i);
d[i] = d[u]+1;
}
}
return d[en];
}
int main()
{
bool f = false;
scanf("%d", &T);
getchar();
while ( T-- ) {
if ( f ) cout << endl;
f = true, id = 0;
//for ( int i = 0; i < N; ++i ) for ( int j = i; j < N; ++j ) g[i][j] = g[j][i] = INF;
memset(g, 0, sizeof(g));
mp.clear();
while ( cin >> s[id] && s[id] != "*" ){
if ( !mp[s[id]] ) {
mp[s[id]] = id;
for ( int i = 0; i < id; ++i )
if ( s[id].size() == s[i].size() ) {
int num = 0;
for ( int k = 0; k < s[id].size() && num < 2; ++k ) if ( s[id][k] != s[i][k] ) num++;
if ( num == 1 ) g[id][i] = g[i][id] = 1;
}
id++;
}
}
//for ( int k = 0; k < id; ++k ) for ( int i = 0; i < id; ++i ) for ( int j = 0; j < id; ++j )
// if ( g[i][k] + g[k][j] < g[i][j] ) g[i][j] = g[i][k] + g[k][j];
char ch;
getchar();
while ( ( ch = getchar() ) != 0xa && ch != EOF ) {
cin >> s1;
s2.clear();
s2 += ch; s2 += s1; cin >> s1;
getchar();
int u = mp[s2], v = mp[s1];
cout << s2 << " " << s1 << " " << bfs( u, v ) << endl;
}
}
}
这道题可以用BFS来解, floyd也没有问题
但是一定要注意,输入单词对的时候,结束符分两种情况,一种是 几组连续输入中间的是空行为结束,最后一组数据是以EOF结束
注:注释掉的 有一部分是floyd算,但是一下正常的是用BFS解题
代码:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <map>
#include <queue>
using namespace std;
const int N = 300;
const int INF = 1000000;
int T, id;
int g
, d
;
map <string, int> mp;
string s1, s2, s
;
int bfs( int st, int en ) {
queue< int > q;
q.push(st);
for ( int i = 0; i <= id; ++i ) d[i] = INF;
d[st] = 0;
while ( !q.empty() ) {
int u = q.front(); q.pop();
if ( u == en ) break;
for ( int i = 0; i < id; ++i ) if ( g[u][i] && d[i] > d[u] + 1 ) {
q.push(i);
d[i] = d[u]+1;
}
}
return d[en];
}
int main()
{
bool f = false;
scanf("%d", &T);
getchar();
while ( T-- ) {
if ( f ) cout << endl;
f = true, id = 0;
//for ( int i = 0; i < N; ++i ) for ( int j = i; j < N; ++j ) g[i][j] = g[j][i] = INF;
memset(g, 0, sizeof(g));
mp.clear();
while ( cin >> s[id] && s[id] != "*" ){
if ( !mp[s[id]] ) {
mp[s[id]] = id;
for ( int i = 0; i < id; ++i )
if ( s[id].size() == s[i].size() ) {
int num = 0;
for ( int k = 0; k < s[id].size() && num < 2; ++k ) if ( s[id][k] != s[i][k] ) num++;
if ( num == 1 ) g[id][i] = g[i][id] = 1;
}
id++;
}
}
//for ( int k = 0; k < id; ++k ) for ( int i = 0; i < id; ++i ) for ( int j = 0; j < id; ++j )
// if ( g[i][k] + g[k][j] < g[i][j] ) g[i][j] = g[i][k] + g[k][j];
char ch;
getchar();
while ( ( ch = getchar() ) != 0xa && ch != EOF ) {
cin >> s1;
s2.clear();
s2 += ch; s2 += s1; cin >> s1;
getchar();
int u = mp[s2], v = mp[s1];
cout << s2 << " " << s1 << " " << bfs( u, v ) << endl;
}
}
}
相关文章推荐
- uva 558(最短路径)
- uva 10985(最短路径)
- UVa - 10278 Fire Station 消防站 图上的多源最短路径
- uva 1416 Warfare And Logistics (最短路径树)
- POJ 2235 Frogger / UVA 534 Frogger /ZOJ 1942 Frogger(图论,最短路径)
- uva117 最短路径
- uva 10269(最短路径)
- The Postal Worker Rings Once(UVA 117)最短路径—SPFA算法+邻接表
- uva 10803(最短路径)
- uva 10801(最短路径)
- UVa10816 - Travel in Desert(二分法+最短路径求法)
- uva 567(最短路径)
- uva 10048(最短路径)
- uva 10986(最短路径)
- UVa10917 - Walk Through the Forest(单源最短路径及动态规划)
- uva11374迪杰斯特拉最短路径+打印
- UVa 10246 Asterix and Obelix(变形的最短路径)
- uva10986(最短路径)
- POJ 1511 Invitation Cards / UVA 721 Invitation Cards / SPOJ Invitation / UVAlive Invitation Cards / SCU 1132 Invitation Cards / ZOJ 2008 Invitation Cards / HDU 1535 (图论,最短路径)
- Uva 11090 最短路径