[bzoj2709][Violet 1]迷宫花园 二分+spfa

2709: [Violet 1]迷宫花园

Time Limit: 5 Sec  Memory Limit: 128 MB
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Description

Input

Output

Sample Input

卡二分的sbt

这是A了的

double l = 0.0, r = 10.0;while( r - l > eps ){double mid = ( l + r ) / 2.0;if( spfa( mid ) ) l = mid + eps;else r = mid - eps;}

WA了的

while( r - l > eps ){double mid = ( l + r ) / 2.0;if( spfa( mid ) ) ans = l = mid;else r = mid;}

再也不做这种sbt了，浪费我时间

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <bitset>#include <queue>#include <ctime>#include <map>#include <set>#define pa pair < int , int >using namespace std;#define eps 1e-10int T, n, dx[4] = { 0, 0, 1, -1 },  dy[4] = { 1, -1, 0, 0 };int r, c, sx, sy, tx, ty;char a[105][105];double Sum, dis[105][105], ans;queue < pa > q; bool inq[105][105];bool spfa( double dang ){memset( dis, 127, sizeof(dis) );memset( inq, 0, sizeof(inq) );while( !q.empty() ) q.pop();dis[sx][sy] = 0; q.push( make_pair( sx, sy ) ); inq[sx][sy] = 1;while( !q.empty() ){int xx = q.front().first, yy = q.front().second; q.pop(); inq[xx][yy] = 0;for( int k = 0; k < 2; k++ ){int xxx = xx + dx[k], yyy = yy + dy[k];if( xxx > 0 && yyy > 0 && xxx <= r && yyy <= c && a[xxx][yyy] != '#' ){if( dis[xxx][yyy] > dis[xx][yy] + 1.0 ){dis[xxx][yyy] = dis[xx][yy] + 1.0;if( !inq[xxx][yyy] ) q.push( make_pair( xxx, yyy ) ), inq[xxx][yyy] = 1;}}}for( int k = 2; k < 4; k++ ){int xxx = xx + dx[k], yyy = yy + dy[k];if( xxx > 0 && yyy > 0 && xxx <= r && yyy <= c && a[xxx][yyy] != '#' ){if( dis[xxx][yyy] > dis[xx][yy] + dang ){dis[xxx][yyy] = dis[xx][yy] + dang;if( !inq[xxx][yyy] ) q.push( make_pair( xxx, yyy ) ), inq[xxx][yyy] = 1;}}}}return dis[tx][ty] < Sum;}int main(){scanf( "%d", &T );while( T-- ){memset( a, 0, sizeof(a) );scanf( "%lf", &Sum ); scanf( "%d%d\n", &r, &c );for( int i = 1; i <= r; i++ ){gets(a[i] + 1);for( int j = 1; j <= c; j++ ){if( a[i][j] == 'S' ) sx = i, sy = j;if( a[i][j] == 'E' ) tx = i, ty = j;}}double l = 0.0, r = 10.0;while( r - l > eps ){double mid = ( l + r ) / 2.0;if( spfa( mid ) ) ans = l = mid;else r = mid;}printf( "%.5lf\n", ans );}return 0;}