POJ 2912 Rochambeau(带权并查集)

这道题纠结了一下午加一晚上，整个人都不好了，而且网上就只有一种解法，估计是题解~

自己想了一种方法无限wa，无奈耐力不够，造数据的能力也不行，不能找到bug，不能像某些偏执狂大神一样为了做出来死磕三天三夜~~~~~~~

最后用了网上的方法1a了，真是无奈。

以后有机会再研究研究。

下面是根据网上的方法写的

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;int father[505],Rank[505],n,m,rec[505];int query(int x) {    if(x != father[x]) {        int per = father[x];        father[x] = query(father[x]);        Rank[x] = (Rank[x] + Rank[per] + 3) % 3;    }    return father[x];}struct Edge {    int x,y;    char c;} edge[2005];void init() {    for(int i = 0;i <= n;i++) {        Rank[i] = 0;        father[i] = i;    }}int main() {    while(~scanf("%d%d",&n,&m)) {        memset(rec,0,sizeof(rec));        for(int i = 0;i < m;i++)            scanf("%d%c%d",&edge[i].x,&edge[i].c,&edge[i].y);        for(int i = 0;i < n;i++) {            init();            for(int j = 0;j < m;j++) {                int x = edge[j].x;                int y = edge[j].y;                if(x == i || y == i)                    continue;                char c = edge[j].c;                if(c == '<')                    swap(x,y);                int a = query(x);                int b = query(y);                if(a == b) {                    if(c == '=' && (Rank[y] - Rank[x] + 3) % 3 != 0) {                        rec[i] = j+1;                        break;                    }                    if(c != '=' && (Rank[y] - Rank[x] + 3) % 3 != 1) {                        rec[i] = j+1;                        break;                    }                }                else {                    father[b] = a;                    if(c == '=')                        Rank[b] = (Rank[x] - Rank[y] + 3) % 3;                    else                         Rank[b] = (Rank[x] - Rank[y] + 4) % 3;                }            }        }        int cnt = 0,k,ans = 0;        for(int i = 0;i < n;i++) {            if(rec[i] == 0) {                k = i;                cnt++;            }            if(rec[i] > ans)                ans = rec[i];        }        /*for(int i = 0;i < n;i++)            printf("%d ",rec[i]);        printf("\n");*/        if(cnt == 0)            printf("Impossible\n");        else if(cnt == 1)            printf("Player %d can be determined to be the judge after %d lines\n",k,ans);        else            printf("Can not determine\n");    }    return 0;}

这种方法难点是求后面的ans，就是最少到那条语句可以判断出谁是裁判。

这里是求有漏洞的语句的最大到哪里，后面的语句都是无用的，也就是只需要前面ans个句子就可以判断

下面是我之前的办法。

我的做法是直接并查集输入数据（i = 1~m）的时候检测漏洞，如果检查到有两个漏洞，并且两个漏洞里出现了同一个人，就说明那个人为备选裁判，再重新枚举一次检查以此人为裁判是否符合题意，最后输出。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int maxn = 5005;int father[maxn],Rank[maxn],n,m;struct Edge {    int x,y;    char c;} edge[20005];int query(int x) {    if(x != father[x]) {        int per = father[x];        father[x] = query(father[x]);        Rank[x] = (Rank[x] + Rank[per] + 3) % 3;    }    return father[x];}int main() {    //freopen("hah.txt","r",stdin);    while(~scanf("%d%d",&n,&m)) {        for(int i = 0;i <= n;i++) {            Rank[i] = 0;            father[i] = i;        }        int x,y,ans = -1,ch,ni,cnt = 0,k,flag = 1;        char c;        for(int i = 0;i < m;i++) {            scanf("%d%c%d",&edge[i].x,&edge[i].c,&edge[i].y);            x = edge[i].x;            y = edge[i].y;            c = edge[i].c;            if(c == '<')                swap(x,y);            int a = query(x);            int b = query(y);            if(a == b && flag && cnt < 2) {                if((Rank[y] - Rank[x] + 3) % 3 != 0  && c == '=')                    cnt++;                else if(c != '=' && (Rank[y] - Rank[x] + 3) % 3 != 1 && (Rank[y] - Rank[x] + 3) % 3 != -2)                    cnt++;                  if(cnt == 1) {                    ch = x;                       ni = y;                }                else if((ch == x && ni == y) || (ch == y && ni == x)) {                    cnt--;                    continue;                }                 //printf("%d-%d %d\n",x,y,cnt);                if(cnt == 2) {                    if(ch == x || ch == y) {                        k = i;                        ans = ch;                    }                    else if(ni == x || ni == y) {                        k = i;                        ans = ni;                    }                    else                        flag = 0;                }                         }            if(a != b && flag && cnt < 2) {                father[b] = a;                if(c == '=')                     Rank[b] = (Rank[x] - Rank[y] + 3) % 3;                else                     Rank[b] = (Rank[x] - Rank[y] + 4) % 3;             }        }        for(int i = 0; i <= n;i++) {             Rank[i] = 0;            father[i] = i;        }        if(ans != -1) {            for(int i = 0;i < m;i++) {                if(edge[i].x == ans || edge[i].y == ans)                    continue;                 int xx = edge[i].x;                int yy = edge[i].y;                if(edge[i].c == '<')                    swap(xx,yy);                int a = query(xx);                int b = query(yy);                char cc = edge[i].c;                //printf("%d %d-%d\n",ans,a,b);                if(a == b) {                    if((Rank[yy] - Rank[xx] + 3 ) % 3 != 0  && cc == '=') {                        flag = 0;                        break;                    }                    if(cc != '=' && (Rank[yy] - Rank[xx] + 3) % 3 != 1 && (Rank[yy] - Rank[xx] + 3) % 3 != -2) {                        flag = 0;                        break;                    }                }                else {                    father[b] = a;                    if(cc == '=')                         Rank[b] = (Rank[xx] - Rank[yy] + 3) % 3;                    else                         Rank[b] = (Rank[xx] - Rank[yy] + 4) % 3;                 }            }        }        /*for(int i = 0;i < n;i++)            printf("%d-",Rank[i]);        printf("\n");*/        if(n == 1 && flag) {            if(cnt == 0)                printf("Player 0 can be determined to be the judge after 0 lines\n");            else                 printf("Impossible\n");        }        else {            if(!flag)                 printf("Impossible\n");            else if(ans != -1 && flag && cnt > 1)                printf("Player %d can be determined to be the judge after %d lines\n",ans,k+1);            else                 printf("Can not determine\n");        }    }}