最短路水题

hdu 1596
求边权乘积的最大值，Dijkstra修改下松弛操作即可

#include
using namespace std;

const int maxn = 1005;
const int INF = 0x3f3f3f3f;

double f[maxn][maxn];
double d[maxn];
bool vis[maxn];

int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= n; ++j){
                scanf("%lf",&f[i][j]);
            }
        }
        int q,S,E;
        scanf("%d",&q);
        while(q--){
            scanf("%d%d",&S,&E);
            memset(d,0,sizeof(d));
            d[S] = 1;
            memset(vis,0,sizeof(vis));
            for(int k = 0; k < n; ++k){
                double maxx = 0; int  maxxx;
                for(int j = 1; j <= n; ++j){
                    if(!vis[j] && d[j] > maxx){
                          maxx = d[maxxx = j];
                    }
                }
                vis[maxxx] = 1;
                for(int j = 1; j <= n; ++j){
                    if(d[j] < d[maxxx] * f[maxxx][j]){
                        d[j] = d[maxxx]*f[maxxx][j];
                    }
                }
            }
            if(d[E]) printf("%.3f\n",d[E]);
            else printf("What a pity!\n");
        }
    }
    return 0;
}

hdu 1548
边权为1的最短路，也可以直接BFS

#include
using namespace std;

const int maxn = 205;
const int INF = 0x3f3f3f3f;

int f[maxn][maxn];
int d[maxn];
bool vis[maxn];

int dijkstra(int s, int t, int n){
    memset(vis,0,sizeof(vis));
    memset(d,0x3f,sizeof(d));
    d[s] = 0;
    for(int i = 1; i <= n; ++i){
        int minx = INF, minxx;
        for(int j = 1; j <= n; ++j){
            if(!vis[j] && d[j] < minx){
                minx = d[minxx = j];
            }
        }
        vis[minxx] = 1;
        for(int j = 1; j <= n; ++j){
            if(d[j] > d[minxx] + f[minxx][j]){
                d[j] = d[minxx] + f[minxx][j];
            }
        }
    }
    return d[t];
}

int main(){
    int n,A,B;
    while(~scanf("%d",&n) && n){
        scanf("%d%d",&A,&B);
        memset(f,0x3f,sizeof(f));
        int x;
        for(int i = 1; i <= n; ++i){
            scanf("%d",&x);
            if(i-x >= 1){
                f[i][i-x] = 1;
            }
            if(i+x <= n){
                f[i][i+x] = 1;
            }
        }
        int ans = dijkstra(A,B,n);
        if(ans < INF) printf("%d\n",ans);
        else printf("-1\n");
    }
    return 0;
}

hdu 2066
建一个新的源点s0，s0到点集S中点的距离都为零

#include
using namespace std;

const int maxn = 1005;
const int INF = 0x3f3f3f3f;
int f[maxn][maxn];
int d[maxn];
bool vis[maxn];
int n = 1000;

int dijkstra(){
    memset(vis,0,sizeof(vis));
    memset(d,0x3f,sizeof(d));
    d[0] = 0;
    for(int i = 0; i <= n; ++i){
        int minx = INF, minxx;
        for(int j = 0; j <= n; ++j){
             if(!vis[j] && d[j] < minx){
                minx = d[minxx = j];
             }
        }
        vis[minxx] = 1;
        for(int j = 0; j <= n; ++j){
            d[j] = min(d[j],d[minxx] + f[minxx][j]);
        }
    }
}

int temp[maxn];

int main(){
    int m,S,T;
    while(~scanf("%d%d%d",&m,&S,&T)){
        memset(f,0x3f,sizeof(f));
        int u,v,w;
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d",&u,&v,&w);
            if(w < f[u][v]) f[u][v] = f[v][u] = w;
        }
        int x;
        for(int i = 0; i < S; ++i){
            scanf("%d",&x);
            f[0][x] = f[x][0] = 0;
        }
        for(int i = 0; i < T; ++i){
            scanf("%d",&temp[i]);
        }
        dijkstra();
        int minx = INF;
        for(int i = 0; i < T; ++i){
            if(d[temp[i]] < minx) minx = d[temp[i]];
        }
        printf("%d\n",minx);
    }
    return 0;
}

hdu 2112
利用map记录点的对应的字符串地名，然后floyd

#include
using namespace std;

const int maxn = 155;
const int INF = 0x3f3f3f3f;
int f[maxn][maxn];
int n,m;

void floyd(){
    for(int k = 0; k < n; ++k){
       for(int i = 0 ; i < n; ++i){
            for(int j = 0; j < n; ++j){
                f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
            }
        }
    }
}

map<string,int>cnt;

int main(){
    while(~scanf("%d",&m)){
        if(m == -1) break;
        cnt.clear();
        n = 0;
        string Start, End;
        cin>>Start>>End;
        cnt[Start] = n++;
        cnt[End] = n++;
        memset(f,0x3f,sizeof(f));
        for(int i = 0; i < m; ++i){
             string s1,s2;
             int w;
             cin>>s1>>s2>>w;
             if(!cnt.count(s1)) cnt[s1] = n++;
             if(!cnt.count(s2)) cnt[s2] = n++;
             if(w < f[cnt[s1]][cnt[s2]]) f[cnt[s1]][cnt[s2]] = f[cnt[s2]][cnt[s1]] = w;
        }
        if(Start == End){
            printf("0\n");
            continue;
        }
        floyd();
        if(f[cnt[Start]][cnt[End]] < INF) printf("%d\n",f[cnt[Start]][cnt[End]]);
        else printf("-1\n");
    }
    return 0;
}