nyoj+三个水杯bfs错了很多小心点+单步调试找错误很好

来源：互联网发布：linux如何自建pdnsd 编辑：程序博客网时间：2024/04/29 23:58
点击打开链接
#include<stdio.h>#include<stdlib.h>#include<string.h>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<math.h>#include<queue>using namespace std;struct Node{    int v1,v2,v3,steps;};queue<Node>Q;int V1=0,V2=0,V3=0,E1=0,E2=0,E3=0,visit[105][105][105];int check(Node temp){    if(temp.v1==E1&&temp.v2==E2&&temp.v3==E3)        return 1;    return 0;}void BFS(){    while(!Q.empty()) Q.pop();    memset(visit,0,sizeof(visit));    Node n,temp,temp1;    n.v1=V1,n.v2=0,n.v3=0,n.steps=0;///变量赋值出错，炸了几个小时    visit[n.v1][n.v2][n.v3]=1;    Q.push(n);    while(!Q.empty()){        temp1=Q.front(); Q.pop();        if(check(temp1)) {            printf("%d\n",temp1.steps);            return;        }        temp=temp1;///这里一定要变量赋值，因为从队列里出来的temp1.分别进行6次变化。        if(temp.v2!=V2&&temp.v1){///pour V1->V2,temp.v1有水且temp.v2未满            if(temp.v1<=V2-temp.v2){///V1 water is not enough                temp.v2+=temp.v1;                temp.v1=0;            }            else{                temp.v1-=(V2-temp.v2);                temp.v2=V2;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;///最少步数嘛，一定要标记防重复            }        }        temp=temp1;        if(temp.v3!=V3&&temp.v1){///pour V1->V3            if(temp.v1<=V3-temp.v3){///V1 water is not enough                temp.v3+=temp.v1;                temp.v1=0;            }            else{                temp.v1-=(V3-temp.v3);                temp.v3=V3;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;            }        }          temp=temp1;          if(temp.v1!=V1&&temp.v2){///pour V2->V1            if(temp.v2<=V1-temp.v1){///V2 water is not enough                temp.v1+=temp.v2;                temp.v2=0;            }            else{                temp.v2-=(V1-temp.v1);                temp.v1=V1;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;            }        }         temp=temp1;         if(temp.v3!=V3&&temp.v2){///pour V2->V3            if(temp.v2<=V3-temp.v3){///V2 water is not enough                temp.v3+=temp.v2;                temp.v2=0;            }            else{                temp.v2-=(V3-temp.v3);                temp.v3=V3;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;            }        }          temp=temp1;         if(temp.v1!=V1&&temp.v3){///pour V3->V1            if(temp.v3<=V1-temp.v1){///V1 water is not enough                temp.v1+=temp.v3;                temp.v3=0;            }            else{                temp.v3-=(V1-temp.v1);                temp.v1=V1;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;            }        }         temp=temp1;         if(temp.v2!=V2&&temp.v3){///pour V3->V2            if(temp.v3<=V2-temp.v2){///V1 water is not enough                temp.v2+=temp.v3;                temp.v3=0;            }            else{                temp.v3-=(V2-temp.v2);                temp.v2=V2;            }            if(!visit[temp.v1][temp.v2][temp.v3]){                temp.steps+=1;                Q.push(temp);                visit[temp.v1][temp.v2][temp.v3]=1;            }        }    }  printf("-1\n");}int main(){   int N=0;   scanf("%d",&N);   while(N--){        scanf("%d%d%d%d%d%d",&V1,&V2,&V3,&E1,&E2,&E3);        BFS();   }   return 0;}
0 0