HDU6071 Lazy Running【最短路】

题意：一共4个点，从2出发，回到2，求大于等于k的最短路

思路：如果存在回到2的路径为k的路，那k+2w的路径一定存在，那只要求d[2][p]的最短路，p为路径长%2w。把这些值加2w加到超过k，取最小值。

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 60005; int dis[5][5],w,inq[5][maxn];ll d[5][maxn];struct data{int p,wei;};void spfa(){queue<data> q;while(!q.empty())q.pop();memset(inq,0,sizeof inq);memset(d,0x3f,sizeof d);d[2][0] = 0;q.push((data){2,0});while(!q.empty()){int x = q.front().p;int we = q.front().wei;q.pop();inq[x][we] = 0;for(int i = -1; i <= 1; i += 2){int y = (x+4+i) % 4;if(y == 0) y = 4;ll z = d[x][we] + dis[x][y];if(d[y][z%w] > z){d[y][z%w] = z;if(!inq[y][z%w]){inq[y][z%w] = 1;q.push((data){y,z%w});}}}}}int main(void){int T,a,b,c,e;ll k;scanf("%d",&T);while(T--){scanf("%lld%d%d%d%d",&k,&a,&b,&c,&e);dis[1][2] = dis[2][1] = a;dis[2][3] = dis[3][2] = b;dis[3][4] = dis[4][3] = c;dis[4][1] = dis[1][4] = e;w = 2 * min(dis[1][2],dis[2][3]);spfa();ll ans = 0x3f3f3f3f3f3f3f3f;for(int i = 0; i < w; i++){ll temp = k - d[2][i];if(temp <= 0)ans = min(ans,d[2][i]);elseans = min(ans,d[2][i] + (temp/w) * w + ( (temp%w)?1:0 ) * w );}printf("%lld\n",ans);}return 0;}