[Offer收割]编程练习赛35

有歧义的号码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int cmp(const void * x, const void * y) {
    //x < y
    return (*((int *)(x))) > (*((int *)(y))) ? 1 : -1;
}
int a[10], b[10];
const int r[10] = { 0, 1, 2, -1, -1, 5, 9, -1, 8, 6 };
int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, d, x, y;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++){
        x = i, d = 0;
        if (x % 10 == 0) continue;
        bool flag = true;
        while (x){
            a[d++] = x % 10;
            x /= 10;
        }
        for (int j = 0; j < d; j++){
            b[j] = r[a[d - j - 1]];
            if (b[j] == -1) flag = false;
        }
        y = 0;
        for (int j = d - 1; j >= 0; j--){
            y = y * 10 + b[j];
        }
        if (y>n || y == i) flag = false;
        if (flag) printf("%d
", i);
    }
    return 0;
}

最短游览路线

算以1为源点的单源最短路，选择能走到1点的点中路径最短的一条加一输出。

没考虑-1的情况，也过了。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int cmp(const void * x, const void * y) {
    //x < y
    return (*((int *)(x))) > (*((int *)(y))) ? 1 : -1;
}
#include <vector>
#include <queue>
using namespace std;
vector<int> G[10005];
queue<int> q;
bool inq[10005], v[10005];
int d[10005];
int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, m, a, b, ans;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++){
        G[i].clear();
        d[i] = 0x3FFFFFFF;
    }
    d[1] = 0;
    memset(v, false, sizeof(v));
    for (int i = 0; i < m; i++){
        scanf("%d%d", &a, &b);
        G[a].push_back(b);
        if (b == 1) v[a] = true;
    }
    memset(inq, false, sizeof(inq));
    while (!q.empty()) q.pop();
    q.push(1);
    inq[1] = true;
    while (!q.empty()){
        int x = q.front();
        q.pop();
        inq[x] = false;
        for (int i = 0; i < G[x].size(); i++){
            int u = G[x][i];
            if (d[u]>d[x] + 1){
                d[u] = d[x] + 1;
                if (!inq[u]){
                    inq[u] = true;
                    q.push(u);
                }
            }
        }
    }
    ans = 0x3FFFFFFF;
    for (int i = 1; i <= n; i++){
        if (v[i] && d[i] < ans) ans = d[i];
    }
    printf("%d
", ans + 1);
    return 0;
}

 重复字符串匹配

缩写命名

二分图匹配，字典中的单词是一个点集，S中的各个字母是一个点集，若最大匹配为|S|输出Yes，否则输出No。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int cmp(const void * x, const void * y) {
    //x < y
    return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
class MaxFlow_SAP {
public:
#define SAP_MAXN 20010
#define SAP_MAXM 880010
#define SAP_INF 0x3FFFFFFF

    struct Node {
        int from, to, next;
        int cap;
    } edge[SAP_MAXM];
    int tol;
    int head[SAP_MAXN];
    int dep[SAP_MAXN];
    int gap[SAP_MAXN];
    int cur[SAP_MAXN];
    int S[SAP_MAXN];
    int que[SAP_MAXN];
    int n;
    MaxFlow_SAP() {
        clear();
    }
    void clear() {
        tol = 0;
        memset(head, -1, sizeof(head));
    }

    void addedge(int u, int v, int w) {
        edge[tol].from = u;
        edge[tol].to = v;
        edge[tol].cap = w;
        edge[tol].next = head[u];
        head[u] = tol++;
        edge[tol].from = v;
        edge[tol].to = u;
        edge[tol].cap = 0;
        edge[tol].next = head[v];
        head[v] = tol++;
    }
    void BFS(int start, int end) {
        memset(dep, -1, sizeof(dep));
        memset(gap, 0, sizeof(gap));
        gap[0] = 1;
        int front, rear;
        front = rear = 0;
        dep[end] = 0;
        que[rear++] = end;
        while (front != rear) {
            int u = que[front++];
            if (front == SAP_MAXN) {
                front = 0;
            }
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].to;
                if (dep[v] != -1) {
                    continue;
                }
                que[rear++] = v;
                if (rear == SAP_MAXN) {
                    rear = 0;
                }
                dep[v] = dep[u] + 1;
                ++gap[dep[v]];
            }
        }
    }
    int SAP(int start, int end) {
        int res = 0;
        BFS(start, end);
        int top = 0;
        memcpy(cur, head, sizeof(head));
        int u = start;
        int i;
        while (dep[start] < n) {
            if (u == end) {
                int temp = SAP_INF;
                int inser;
                for (i = 0; i < top; i++)
                    if (temp > edge[S[i]].cap) {
                        temp = edge[S[i]].cap;
                        inser = i;
                    }
                for (i = 0; i < top; i++) {
                    edge[S[i]].cap -= temp;
                    edge[S[i] ^ 1].cap += temp;
                }
                res += temp;
                top = inser;
                u = edge[S[top]].from;
            }
            if (u != end && gap[dep[u] - 1] == 0) {
                break;
            }
            for (i = cur[u]; i != -1; i = edge[i].next)
                if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1) {
                    break;
                }
            if (i != -1) {
                cur[u] = i;
                S[top++] = i;
                u = edge[i].to;
            } else {
                int min = n;
                for (i = head[u]; i != -1; i = edge[i].next) {
                    if (edge[i].cap == 0) {
                        continue;
                    }
                    if (min > dep[edge[i].to]) {
                        min = dep[edge[i].to];
                        cur[u] = i;
                    }
                }
                --gap[dep[u]];
                dep[u] = min + 1;
                ++gap[dep[u]];
                if (u != start) {
                    u = edge[S[--top]].from;
                }
            }
        }
        return res;
    }
};
char str[105], dict[105][105];
MaxFlow_SAP sap;
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n, s, t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        scanf("%s", str);
        s = strlen(str);
        for (int i = 0; i < n; i++) {
            scanf("%s", dict[i]);
        }
        sap.clear(), sap.n = n + s + 2;
        for (int i = 2; i <= n + 1; i++) {
            sap.addedge(1, i, 1);
        }
        for (int i = n + 2; i <= n + s + 1; i++) {
            sap.addedge(i, n + s + 2, 1);
        }
        for (int i = 0; i < n; i++) {
            int m = strlen(dict[i]);
            for (int j = 0; j < m; j++) {
                for (int k = 0; k < s; k++) {
                    if (dict[i][j] == str[k]) {
                        sap.addedge(i + 2, n  + k + 2, 1);
                    }
                }
            }
        }
        int maxflow = sap.SAP(1, n + s + 2);
        if (maxflow == s) printf("Yes
");
        else printf("No
");
        //1 2~n+1 n+2~n+s+1 n+s+2
    }
    return 0;
}