fill与memset的区别介绍

例一

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=500;
const int INF=1000000000;
bool isin[maxn]={false};
int G[maxn][maxn];
int path[maxn],rescue[maxn],num[maxn];
int weight[maxn];
int citynum,roadnum,begins,e;void Dijisktra(int a){fill(path,path+maxn,INF);//记录最短距离memset(rescue,0,sizeof(rescue));//记录点权memset(num,0,sizeof(num));//记录最短路径的条数path[a]=0;rescue[a]=weight[a];num[a]=1;for(int i=0;i<citynum;i++){int pi=-1,pv=INF;for(int j=0;j<citynum;j++){//找最短距离的结点if(isin[j]==false&&path[j]<pv){pi=j;pv=path[j];}}if(pi==-1) return;isin[pi]=true;for(int k=0;k<citynum;k++){if(isin[k]==false&&G[pi][k]!=INF){if(G[pi][k]+path[pi]<path[k]){path[k]=G[pi][k]+path[pi];num[k]=num[pi];//更新最短路径数：不相同就覆盖rescue[k]=rescue[pi]+weight[k];}else if(G[pi][k]+path[pi]==path[k]){if(rescue[pi]+weight[k]>rescue[k])rescue[k]=rescue[pi]+weight[k];//存大值num[k]+=num[pi];//最短路径条数之和：相同累加}}}}
}
int main(){int v1,v2;//顶点及边权-距离fill(G[0],G[0]+maxn*maxn,INF);cin>>citynum>>roadnum>>begins>>e;for(int i=0;i<citynum;i++){cin>>weight[i];//记录点权-救援小组数目}for(int j=0;j<roadnum;j++){cin>>v1>>v2>>G[v1][v2];//建立无向图G[v2][v1]=G[v1][v2];}Dijisktra(begins);cout<<num[e]<<" "<<rescue[e];return 0;
}

例二

#include <iostream>
using namespace std;
const int maxn=100;
const int INF=1000000000;
bool isin[maxn]={false};
int G[maxn][maxn],expense[maxn][maxn];
int path[maxn],cost[maxn],pre[maxn];
int citynum,roadnum,b,e;void Dijisktra(int a){//求最短路径fill(path,path+maxn,INF);fill(cost,cost+maxn,INF);path[a]=0;cost[a]=0;for(int i=0;i<citynum;i++) pre[i]=i;for(int i=0;i<citynum;i++){int m=-1,mv=INF;for(int j=0;j<citynum;j++){if(isin[j]==false&&path[j]<mv){m=j;mv=path[j];}}if(m==-1) return;isin[m]=true;for(int k=0;k<citynum;k++){if(isin[k]==false&&G[m][k]!=INF){if(G[m][k]+path[m]<path[k]){path[k]=G[m][k]+path[m];cost[k]=expense[m][k]+cost[m];pre[k]=m;}else if(G[m][k]+path[m]==path[k]){if(cost[k]>expense[m][k]+cost[m])cost[k]=expense[m][k]+cost[m];pre[k]=m;}}}}
}
void DFSprint(int now){//打印if(now==b){cout<<now<<" ";return;}DFSprint(pre[now]);cout<<now<<" ";
}
int main(){int v1,v2;fill(G[0],G[0]+maxn*maxn,INF);fill(expense[0],expense[0]+maxn*maxn,INF);cin>>citynum>>roadnum>>b>>e;for(int i=0;i<roadnum;i++){cin>>v1>>v2>>G[v1][v2]>>expense[v1][v2];G[v2][v1]=G[v1][v2];expense[v2][v1]=expense[v1][v2];}Dijisktra(b);DFSprint(e);cout<<path[e]<<" "<<cost[e]<<endl;return 0;
}

拓展

用迪杰斯特拉+DFS求最短路径的方法
关键代码：

const int maxn=100;
const int INF=10000000000;
bool isin[maxn]={false};
int G[maxn][maxn],num;
int path[maxn],w[maxn];
vector<int> pre[maxn];//记录最短路径的前驱：考虑会有多个
vector<int> minPath,temPath;//只记录最优或当前路径void Dijisktra(int a){fill(path,path+maxn,INF);path[a]=0;for(int i=0;i<num;i++){int m=-1,mv=INF;for(int j=0;j<num;j++){if(isin[j]==false&&path[j]<mv){m=j;mv=path[j];}}if(m==-1) return;isin[m]=true;for(int k=0;k<num;k++){if(isin[k]==false&&G[m][k]!=INF){if(G[m][k]+path[m]<path[k]){path[k]=G[m][k]+path[m];//找到更优路径，清空，装最短的pre[k].clear();//m结点加入k结点的前驱列表中，即为pre[k][i]==m;pre[k].push_back(m);}else if(G[m][k]+path[m]==path[k]){
//此时有多条最短路径，即存在多个前驱结点，直接加入即可pre[k].push_back(m);}}}}
}void DFSprint(int now,int begins){int optValue=0;if(now==begins){temPath.push_back(begins);if(value>optValue){//更新最优optValue=value;minPath=temPath;}temPath.pop_back();//弹出第一个return;}temPath.push_back(now);for(int i=0;i<pre[now].size();i++){//遍历当前结点的前驱结点DFSprint(pre[now][i]);//不断递归now结点的前驱列表}temPath.pop_back();//依次弹出第二个.....
}
//计算边权和
int edge=0;
for(int i=temPath.size()-1;i>0;i--){int now=temPath[i],next=temPath[i-1];edge+=G[now][next];//计算边权
}
//计算点权和
int weight=0;
for(int i=temPath.size()-1;i>0;i--){int now=temPath[i];//当前结点下标weight+=w[now];
}

例二：Dijiskatra+DFS

#include <iostream>
#include <vector>
using namespace std;
const int maxn=100;
const int INF=100000000;
bool isin[maxn]={false};
int G[maxn][maxn],expense[maxn][maxn];
int path[maxn],minValue=INF;
vector<int> pre[maxn],temPath,minPath;
int citynum,roadnum,b,e;void Dijisktra(int a){fill(path,path+maxn,INF);path[a]=0;for(int i=0;i<citynum;i++){int m=-1,mv=INF;for(int j=0;j<citynum;j++){if(isin[j]==false&&path[j]<mv){m=j;mv=path[j];}}if(m==-1) return;isin[m]=true;for(int k=0;k<citynum;k++){if(isin[k]==false&&G[m][k]!=INF){if(path[m]+G[m][k]<path[k]){path[k]=path[m]+G[m][k];pre[k].clear();pre[k].push_back(m);}else if(path[m]+G[m][k]==path[k]){pre[k].push_back(m);}}}}
}
void DFSprint(int now){int tempValue=0;if(now==b){temPath.push_back(now);for(int i=temPath.size()-1;i>0;i--){int v1=temPath[i],v2=temPath[i-1];tempValue+=expense[v1][v2];}if(tempValue<minValue){minValue=tempValue;minPath=temPath;}temPath.pop_back();//出队}temPath.push_back(now);for(int i=0;i<pre[now].size();i++){DFSprint(pre[now][i]);}temPath.pop_back();
}
int main(){int v1,v2;fill(G[0],G[0]+maxn*maxn,INF);fill(expense[0],expense[0]+maxn*maxn,INF);cin>>citynum>>roadnum>>b>>e;for(int i=0;i<roadnum;i++){cin>>v1>>v2>>G[v1][v2]>>expense[v1][v2];G[v2][v1]=G[v1][v2];expense[v2][v1]=expense[v1][v2];}Dijisktra(b);DFSprint(e);for(int i=minPath.size()-1;i>=0;i--)cout<<minPath[i]<<" ";cout<<path[e]<<" "<<minValue<<endl;return 0;
}