UVA 515 King

題目鏈接：http://acm.hust.edu.cn/vjudge/problem/25023

題意：有n個數(a1~an)，有m個關係：給出區間[L,R]和k，∑ai < k 或者 ∑ai > k。然後判斷是否存在這樣一個序列。

思路：差分約束模板題，需要把公式變一下型，構建n+1個點Si(0<=i<=n)表示前i個數的前綴和一個源點。然後關係就變成了Si - Sj > k通過兩邊取反或+1的形式把關係不等式變形成Si-Sj<=k，然後直接套模板。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007
const int maxn = 1009;

struct node
{
    int to,w,next;
}edge[maxn];
int head[maxn];
int tot = 0;
int n,m;
void addedge(int x,int y,int z)
{
    tot++;
    edge[tot].w = z , edge[tot].to = y , edge[tot].next = head[x];
    head[x] = tot;
}
int dis[maxn];

bool spfa( int st  )
{
	bool vis[maxn];
	int outque[maxn];
	memset(vis,0,sizeof(vis));
	memset(outque,0,sizeof(outque));
	for( int i = 0; i <=n+1; i++ ) dis[i] = inf;//初始化距離
	queue<int> q;
	q.push(st) , vis[st] = true , dis[st] = 0;//初始狀態
	while( !q.empty() )
	{
		int u = q.front();
		q.pop();
		outque[u]++ , vis[u] = false;
		if ( outque[u] > n+2 ) return false; //有負環 無最短路

		for( int k = head[u]; k != -1; k = edge[k].next )
		{
			int v = edge[k].to;
			if ( dis[v] > dis[u] + edge[k].w )
			{
				dis[v] = dis[u] + edge[k].w;
				if( !vis[v] )	vis[v] = true , q.push( v );
			}
		}
	}
	return true;
}

char str[maxn];

void init()
{
    int x,y,k;
    cin>>m;
    Clean(head,-1) , tot = 0;
    rep(i,1,m)
    {
        scanf("%d %d %s %d",&x,&y,str,&k);
        if ( str[0] == 'g' )
            addedge(x+y,x-1,-k-1);
        else
            addedge(x-1,x+y,k-1);
    }
    rep(i,0,n) addedge(n+1,i,0);
    if ( spfa(n+1) )
        puts("lamentable kingdom");
    else
        puts("successful conspiracy");
}


int main()
{
    while( cin>>n )
    {
        if ( !n ) break;
        init();
    }
    return 0;
}