HDU 5336 (2015 + school 4)-xyz and Drops (BFS)

Title Address: HDU 5336
Test instructions: There is an R row in column C, giving the size of the water droplets in n squares. The time limit T is given, which causes the water droplets to burst from the (Sx,sy) position, and when the fly-out waters meet the static water in the lattice, they converge when the aggregate droplet size >4. Ask about the water droplets in the N-position lattice given at t when the output is not burst: 1 The size of the droplets in the lattice. Otherwise the output: 0 blasting time.

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;intmp[ the][ the];inttime[ the][ the];intjx[]= {0,1,0,-1};intjy[]= {1,0,-1,0};intR,c,n,t;intSx,sy;structDrop {intx, y;intTimeintPapa;} F1,F2;voidBFsintXxintYY) { queue<drop>Q    F1.x=xx;    F1.y=yy; F1.time=0; f1.papa=-1; Q.push (F1); while(!q.empty ())     {F1=q.front (); Q.pop ();if(f1.papa>=0) {F2.x=f1.x+jx[f1.papa];            F2.y=f1.y+jy[f1.papa]; f2.time=f1.time+1;if(f2.x>=1&&f2.x<=r&&f2.y>=1&&f2.y<=c&&f2.time<=t) {if(Mp[f2.x][f2.y]) {mp[f2.x][f2.y]++;if(mp[f2.x][f2.y]==5) {f2.papa=-1;                        Q.push (F2); }                    }Else{F2.papa=f1.papa;                    Q.push (F2); }                }        }Else{time[f1.x][f1.y]=f1.time; mp[f1.x][f1.y]=0; for(intI=0; i<4;            i++) {F2.x=f1.x+jx[i];            F2.y=f1.y+jy[i]; f2.time=f1.time+1;if(f2.x>=1&&f2.x<=r&&f2.y>=1&&f2.y<=c&&f2.time<=t) {if(Mp[f2.x][f2.y]) {mp[f2.x][f2.y]++;if(mp[f2.x][f2.y]==5) {f2.papa=-1;                        Q.push (F2); }                    }Else{f2.papa=i;                    Q.push (F2); }                }            }        }    }}intMain () {intV Drop node[ the]; while(~scanf("%d %d%d%d", &r,&c,&n,&t)) {memset(MP,0,sizeof(MP)); for(intI=0; i<n;i++) {scanf(" %d%d%d", &sx,&sy,&v);            NODE[I].X=SX;            Node[i].y=sy;        Mp[sx][sy]+=v; }scanf("%d%d", &sx,&sy); BFS (Sx,sy); for(intI=0; i<n;i++)if(Mp[node[i].x][node[i].y])printf("1%d\n", Mp[node[i].x][node[i].y]);Else                printf("0%d\n", Time[node[i].x][node[i].y]); }}/* Arrive at the same water droplet at the same time (so the code is ==5 instead of >4) 4 4 5 1001 1 2 to 3 4 3

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HDU 5336 (2015 + school 4)-xyz and Drops (BFS)