ZOJ3732 Graph Reconstruction（图的构造）

来源：互联网发布：基金南方大数据300 编辑：程序博客网时间：2024/05/31 18:43

主要坑在了多解时，假设最多的度数是 d 则 a[d+2] == a[d+1] 时，d+1 和 d+2 是可以互换的。但是互换只限于发现这对点之后加到图里的边，开始全部互换了错了很多次。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 101;struct node {    int num, deg;    bool operator<(const node &b)const {return deg > b.deg;}}a[maxn];int u[maxn*maxn], v[maxn*maxn];int main(){    int n;    while (~scanf("%d", &n)) {        memset(u, 0, sizeof(u));        memset(v, 0, sizeof(v));        for (int i = 0; i < maxn; i++)            a[i].num = a[i].deg = 0;        bool ok = true, uni = true;        int x, y, cnt = 0, sum = 0, bcnt = 0;        for (int i = 0; i < n; i++) {            a[i].num = i+1;            scanf("%d", &a[i].deg);            sum += a[i].deg;        }        if (sum & 1) {            puts("IMPOSSIBLE");            continue;        }        for (int i = 0; ok && i < n; i++) {            sort(a+i, a+n);            if (a[i].deg == 0) break;            for (int j = 1; j <= a[i].deg; j++) {                if (j == a[i].deg && i+j+1 < n && a[i+j].deg == a[i+j+1].deg) {                    uni = false;                    x = a[i+j].num, y = a[i+j+1].num;                    bcnt = cnt;                }                if (i+j >= n || --a[i+j].deg < 0) {                    ok = false;                    break;                }                u[cnt] = a[i].num;                v[cnt] = a[i+j].num;                cnt++;            }        }        if (!ok)            puts("IMPOSSIBLE");        else if (uni) {            printf("UNIQUE\n%d %d\n", n, cnt);            for (int i = 0; i < cnt; i++)                printf("%d%c", u[i], i == cnt-1 ? '\n' : ' ');            for (int i = 0; i < cnt; i++)                printf("%d%c", v[i], i == cnt-1 ? '\n' : ' ');        } else {            printf("MULTIPLE\n%d %d\n", n, cnt);            for (int i = 0; i < cnt; i++)                printf("%d%c", u[i], i == cnt-1 ? '\n' : ' ');            for (int i = 0; i < cnt; i++)                printf("%d%c", v[i], i == cnt-1 ? '\n' : ' ');            printf("%d %d\n", n, cnt);            for (int i = 0; i < cnt; i++) {                if (i >= bcnt) {                    if (u[i] == x) u[i] = y;                    else if (u[i] == y) u[i] = x;                }                printf("%d%c", u[i], i == cnt-1 ? '\n' : ' ');            }            for (int i = 0; i < cnt; i++) {                if (i >= bcnt) {                    if (v[i] == x) v[i] = y;                    else if (v[i] == y) v[i] = x;                }                printf("%d%c", v[i], i == cnt-1 ? '\n' : ' ');            }        }    }    return 0;}

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