标签：10 虐到 12 Saving int James 30 crocodile data
题面：
This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Assume that the lake is a 100 by 100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him a shortest path to reach one of the banks. The length of a path is the number of jumps that James has to make.
Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the ( location of a crocodile. Note that no two crocodiles are staying at the same position.
Output Specification:
For each test case, if James can escape, output in one line the minimum number of jumps he must make. Then starting from the next line, output the position ( of each crocodile on the path, each pair in one line, from the island to the bank. If it is impossible for James to escape that way, simply give him 0 as the number of jumps. If there are many shortest paths, just output the one with the minimum first jump, which is guaranteed to be unique.
Sample Input 1:
17 15
10 -21
10 21
-40 10
30 -50
20 40
35 10
0 -10
-25 22
40 -40
-30 30
-10 22
0 11
25 21
25 10
10 10
10 35
-30 10Sample Output 1:
4
0 11
10 21
10 35Sample Input 2:
4 13
-12 12
12 12
-12 -12
12 -12Sample Output 2:
0#include <iostream>
#include <queue>
#include <stack>
using namespace std;
typedef struct{
int x,y;
bool ifNearBank;
bool ifConnected;
bool ifVisited;
}crocodile;
crocodile data[120];
int n;
int d;
int times=0;
int minTimes=20000;
int path[120];
int minpath[120];
int minStartPointList[120];
int stNum=0;
int bPath[120];//记录上一个是谁
int bLong[120];//记录终点离原点的距离
bool connectable(int a, int b){
int dis = (data[a].x - data[b].x)*(data[a].x - data[b].x) + (data[a].y - data[b].y)*(data[a].y - data[b].y);
if(dis<=d*d) return true;
else return false;
}
void copyTime(){
for(int i=0; i<minTimes; i++)
minpath[i] = path[i];
}
void dfs(int pos, bool &flag){
data[pos].ifVisited = true;
path[times] = pos;
times++;
if(data[pos].ifNearBank){
flag = true;
if(times<minTimes){
minTimes = times;
copyTime();
}
}
for(int i=0; i<n; i++){
if(!data[i].ifVisited && connectable(i, pos))
dfs(i, flag);
}
times--;
}
void bfs(int pos, bool &flag){
queue<int> q;
q.push(pos);
data[pos].ifVisited = true;
if(data[pos].ifNearBank){
flag = true;
return;
}
while(!q.empty()){
pos = q.front();
q.pop();
for(int i=0; i<n; i++){
if(!data[i].ifVisited && connectable(i, pos)){
data[i].ifVisited = true;
bPath[i] = pos;
if(data[i].ifNearBank){
flag = true;
}
q.push(i);
}
}
}
}
void calHowLong(){
for(int i=0; i<n; i++){
if(data[i].ifNearBank&&data[i].ifVisited){
int j=i;
int k=1;
while(bPath[j]!=j){
j=bPath[j];
k++;
}
bLong[i] = k;
if(minTimes>k) minTimes=k;
}
}
}
int calStartPoint(int p){
while(bPath[p]!=p){
p=bPath[p];
}return p;
}
int cd(int i){
return data[i].x*data[i].x+data[i].y*data[i].y;
}
int main()
{
scanf("%d%d", &n, &d);
if(d>=42.5){
printf("1\n");
return 0;
}
for(int i=0; i<n; i++){
scanf("%d%d", &data[i].x, &data[i].y);
if(cd(i)<=56.25||data[i].x>=50||data[i].x<=-50||data[i].y>=50||data[i].y<=-50){
n--;
i--;
continue;
}
if(cd(i)<=(7.5+d)*(7.5+d)){
data[i].ifConnected = true;
bPath[i]=i;
minStartPointList[stNum] = i;
for(int j=stNum; j>0; j--){
if(cd(minStartPointList[j]) < cd(minStartPointList[j-1])){
int temp;
temp = minStartPointList[j]; minStartPointList[j] = minStartPointList[j-1]; minStartPointList[j-1] = temp;
}
}
stNum++;
}
else
data[i].ifConnected = false;
if(data[i].x+d>=50||data[i].x-d<=-50||data[i].y+d>=50||data[i].y-d<=-50){
data[i].ifNearBank = true;
bLong[i]=20000;
}
else
data[i].ifNearBank = false;
data[i].ifVisited = false;
}
bool ifEscape = false;
for(int i=0; i<stNum; i++){
times = 0;
bfs(minStartPointList[i], ifEscape);
}
if(ifEscape){
// calStartPoint();
calHowLong();
int finalPoint=-1;
for(int i=0; i<n; i++){
if(minTimes == bLong[i]){
if(finalPoint>=0){
if(cd(calStartPoint(i))<cd(calStartPoint(finalPoint))){
finalPoint = i;
}
}else finalPoint = i;
}
}
stack<int> fuck;
fuck.push(finalPoint);
while(bPath[finalPoint]!=finalPoint){
finalPoint=bPath[finalPoint];
fuck.push(finalPoint);
}
printf("%d\n", minTimes+1);
while(!fuck.empty()){
printf("%d %d\n", data[fuck.top()].x, data[fuck.top()].y);
fuck.pop();
}
}
else printf("0\n");
return 0;
}
题解心得：代码写的很乱，暂时不想调整了，A了就行。 踩了的坑点： 1.之前的easy version用了dfs，但是这道题是不能用dfs的！！！dfs一条路是走到黑，有可能找不到那条最短的路径（可能把能走最短的路径绕远走掉，导致最短路径走不了）。 2.bfs记录路径，我的方法是记录每一个顶点的父顶点，最后通过可以一步到达岸边的那些顶点回溯，将回溯的过程入栈再出栈即为所求的路径。 3.暂时先写这么多。这道题写了我一个下午加半个晚上，现在脑壳疼痛，出去跑会步。。
标签：10,虐到,12,Saving,int,James,30,crocodile,data 来源： https://www.cnblogs.com/wstnl/p/14801967.html
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