现有N个未知数A[1],A[2],…A[N],以及M个方程,每个方程都是形如A[s]+A[s+1]+A[s+2]+…A[t-1]+A[t]=c。现在求解这个方程组。
输入的第一行为两个整数N和M(1<=N,M<=100000)。接下来的M行每行三个整数s,t,c(1 <= s, t <= N, 0 <= c < 10^9)。
对于输入的每个方程,若该方程与前面的方程矛盾,则输出"Error!"并忽略这个方程,否则输出"Accepted!"。之后对于每个变量,若能求出这个变量的值,则输出这个变量,否则输出"Unknown!"。
3 5 1 3 3 1 3 2 2 3 2 1 1 0 1 1 1
Accepted! Error! Accepted! Error! Accepted! 1 Unknown! Unknown!
// whn6325689 // Mr.Phoebe // http://blog.csdn.net/u013007900 #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #include <fstream> #include <cassert> #include <cstdio> #include <bitset> #include <vector> #include <deque> #include <queue> #include <stack> #include <ctime> #include <set> #include <map> #include <cmath> #include <functional> #include <numeric> #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; typedef long long ll; typedef long double ld; typedef pair<ll, ll> pll; typedef complex<ld> point; typedef pair<int, int> pii; typedef pair<pii, int> piii; typedef vector<int> vi; #define CLR(x,y) memset(x,y,sizeof(x)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define lowbit(x) (x&(-x)) #define MID(x,y) (x+((y-x)>>1)) #define eps 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LLINF 1LL<<62 template<class T> inline bool read(T &n) { T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } template <class T> inline void write(T n) { if(n < 0) { putchar('-'); n = -n; } int len = 0,data[20]; while(n) { data[len++] = n%10; n /= 10; } if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } //----------------------------------- const int MAXN=100010; struct DisjointSet { int fa[MAXN]; ll sum[MAXN]; void init(int n) { for(int i=0;i<=n;i++) fa[i]=i; CLR(sum,0); } int findfa(int x) { if(fa[x]==x) return x; int xx=findfa(fa[x]); sum[x]+=sum[fa[x]]; return fa[x]=xx; } bool merge(int x,int y,int num) { int xx=findfa(x); int yy=findfa(y); if(xx==yy) { return sum[x]-sum[y]==num; } if(yy>xx) { fa[xx]=yy; sum[xx]=sum[y]+num-sum[x]; } else { fa[yy]=xx; sum[yy]=sum[x]-num-sum[y]; } return true; } }g; int main() { int n,m,a,b,c; scanf("%d %d",&n,&m); g.init(n); for(int i=0;i<m;i++) { scanf("%d %d %d",&a,&b,&c); b++; if(g.merge(a,b,c)) puts("Accepted!"); else puts("Error!"); } for(int i=1;i<=n;i++) { int xx=g.findfa(i),yy=g.findfa(i+1); if(xx==yy) write(g.sum[i]-g.sum[i+1]),putchar('\n'); else puts("Unknown!"); } return 0; }
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6637 | Accepted: 2580 |
Description
Input
Output
Sample Input
10 5 1 2 even 3 4 odd 5 6 even 1 6 even 7 10 odd
Sample Output
3
// whn6325689 // Mr.Phoebe // http://blog.csdn.net/u013007900 #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #include <fstream> #include <cassert> #include <cstdio> #include <bitset> #include <vector> #include <deque> #include <queue> #include <stack> #include <ctime> #include <set> #include <map> #include <cmath> #include <functional> #include <numeric> #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; typedef long long ll; typedef long double ld; typedef pair<ll, ll> pll; typedef complex<ld> point; typedef pair<int, int> pii; typedef pair<pii, int> piii; typedef vector<int> vi; #define CLR(x,y) memset(x,y,sizeof(x)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define lowbit(x) (x&(-x)) #define MID(x,y) (x+((y-x)>>1)) #define eps 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LLINF 1LL<<62 template<class T> inline bool read(T &n) { T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } template <class T> inline void write(T n) { if(n < 0) { putchar('-'); n = -n; } int len = 0,data[20]; while(n) { data[len++] = n%10; n /= 10; } if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } //----------------------------------- #define ls i<<1 #define rs i<<1|1 const int MAXN=100010; struct DisjointSet { int fa[MAXN]; int sum[MAXN]; void init(int n) { for(int i=0;i<=n;i++) fa[i]=i; CLR(sum,0); } int findfa(int x) { if(fa[x]==x) return x; int xx=findfa(fa[x]); sum[x]+=sum[fa[x]]; return fa[x]=xx; } bool merge(int x,int y,int num) { int xx=findfa(x); int yy=findfa(y); if(xx!=yy) { fa[xx]=yy;sum[xx]=sum[y]+num-sum[x]; } else if((abs(sum[x]-sum[y])&1)!=num) return false; return true; } }g; char str[11]; int a[MAXN],b[MAXN],c[MAXN]; int main() { int n,m,ans=0; while(scanf("%d",&n)!=EOF&&~n) { scanf("%d",&m); for(int i=0;i<m;i++) { scanf("%d %d %s",&a[ls],&a[rs],str); c[i]=str[0]=='o'; a[rs]++; b[ls]=a[ls]; b[rs]=a[rs]; } sort(b,b+2*m); int size=unique(b,b+2*m)-b; g.init(2*m); for(int i=0;i<2*m;i++) a[i]=upper_bound(b,b+size,a[i])-b; ans=m; for(int i=0;i<m;i++) { int x=a[ls]; int y=a[rs]; if(!g.merge(x,y,c[i])) { ans=i; break; } } write(ans),putchar('\n'); } return 0; }